Chapter 5: Continuity and Differentiability

A function $$f$$ is continuous at a point $$x = c$$ if $$\lim_{x \to c} f(x) = f(c)$$.

Here $$f(x) = 2x + 3$$ and $$c = 1$$.

Value of the function at $$x = 1$$:

$$f(1) = 2(1) + 3 = 5$$

Limit of the function as $$x \to 1$$:

$$\lim_{x \to 1} f(x) = \lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5$$

Since $$\lim_{x \to 1} f(x) = 5 = f(1)$$, the function $$f$$ is continuous at $$x = 1$$.

A function $$f$$ is continuous at $$x = c$$ if $$\lim_{x \to c} f(x) = f(c)$$.

Value at $$x = 0$$: $$f(0) = 0^2 = 0$$.

Limit as $$x \to 0$$: $$\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 = 0^2 = 0$$.

Since $$\lim_{x \to 0} f(x) = 0 = f(0)$$, the function $$f$$ is continuous at $$x = 0$$.

Here $$f(x) = |x|$$ and $$c = 0$$. Recall $$|x| = x$$ for $$x \ge 0$$ and $$|x| = -x$$ for $$x < 0$$.

Value at $$x = 0$$: $$f(0) = |0| = 0$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$$.

Right hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$$.

Since the left and right hand limits are equal, $$\lim_{x \to 0} f(x) = 0$$, and this equals $$f(0) = 0$$.

Hence $$f$$ is continuous at $$x = 0$$.

From the definition, $$f(0) = 1$$.

Limit as $$x \to 0$$: For $$x \ne 0$$, $$f(x) = x^3 + 3$$, so

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^3 + 3) = 0 + 3 = 3.$$

Since $$\lim_{x \to 0} f(x) = 3$$ but $$f(0) = 1$$, we have $$\lim_{x \to 0} f(x) \ne f(0)$$.

Therefore $$f$$ is not continuous at $$x = 0$$.

Let $$f(x) = k$$ for all $$x \in \mathbf{R}$$, where $$k$$ is a constant. Let $$c$$ be any real number.

Value at $$x = c$$: $$f(c) = k$$.

Limit as $$x \to c$$: $$\lim_{x \to c} f(x) = \lim_{x \to c} k = k$$.

Since $$\lim_{x \to c} f(x) = k = f(c)$$ for every real number $$c$$, the constant function $$f$$ is continuous at every real number.

A function $$f$$ is continuous at a point $$x = c$$ if $$\lim_{x \to c} f(x) = f(c)$$.

Here $$f(x) = x$$. Let $$c$$ be any arbitrary real number.

Value at $$x = c$$: $$f(c) = c$$.

Limit as $$x \to c$$: $$\lim_{x \to c} f(x) = \lim_{x \to c} x = c$$.

Thus $$\lim_{x \to c} f(x) = c = f(c)$$, so $$f$$ is continuous at $$x = c$$.

Since $$c$$ was an arbitrarily chosen real number, the identity function $$f(x) = x$$ is continuous at every real number.

The function $$f(x) = |x|$$ can be written as $$f(x) = \begin{cases} -x, & \text{if } x \lt 0 \\ x, & \text{if } x \ge 0 \end{cases}$$

Case 1: $$c \lt 0$$. Here $$f(c) = -c$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c} (-x) = -c = f(c)$$. So $$f$$ is continuous at every $$c \lt 0$$.

Case 2: $$c \gt 0$$. Here $$f(c) = c$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c} x = c = f(c)$$. So $$f$$ is continuous at every $$c \gt 0$$.

Case 3: $$c = 0$$. Here $$f(0) = 0$$. The left hand limit is $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$$ and the right hand limit is $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$$, so $$\lim_{x \to 0} f(x) = 0 = f(0)$$. So $$f$$ is continuous at $$x = 0$$.

Since $$f$$ is continuous at every real number, $$f(x) = |x|$$ is a continuous function.

$$f(x) = x^3 + x^2 - 1$$ is a polynomial function. Let $$c$$ be any real number.

Value at $$x = c$$: $$f(c) = c^3 + c^2 - 1$$.

Limit as $$x \to c$$: Using the algebra of limits,

$$\lim_{x \to c} f(x) = \lim_{x \to c} (x^3 + x^2 - 1) = c^3 + c^2 - 1$$

Thus $$\lim_{x \to c} f(x) = c^3 + c^2 - 1 = f(c)$$.

Since this holds for every real number $$c$$, the function $$f$$ is continuous at every real number, i.e. $$f$$ is a continuous function.

The function $$f(x) = \dfrac{1}{x}$$ is defined for all real numbers except $$x = 0$$; its domain is $$\mathbf{R} - \{0\}$$.

Let $$c$$ be any non-zero real number. Then

$$\lim_{x \to c} f(x) = \lim_{x \to c} \dfrac{1}{x} = \dfrac{1}{c}$$

and $$f(c) = \dfrac{1}{c}$$.

Hence $$\lim_{x \to c} f(x) = \dfrac{1}{c} = f(c)$$, so $$f$$ is continuous at every point $$c$$ of its domain.

Therefore $$f$$ is a continuous function. Note that $$x = 0$$ does not belong to the domain of $$f$$, so continuity is not discussed there.

At any point $$c \lt 1$$: Here $$c \le 1$$, so $$f(c) = c + 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(x + 2) = c + 2 = f(c)$$. So $$f$$ is continuous at every $$c \lt 1$$.

At any point $$c \gt 1$$: Here $$f(c) = c - 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(x - 2) = c - 2 = f(c)$$. So $$f$$ is continuous at every $$c \gt 1$$.

At $$c = 1$$: Since $$1 \le 1$$, $$f(1) = 1 + 2 = 3$$.

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x + 2) = 3$$.

Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+}(x - 2) = -1$$.

Since the left and right hand limits are not equal, $$\lim_{x \to 1} f(x)$$ does not exist, so $$f$$ is not continuous at $$x = 1$$.

Hence $$f$$ is continuous at every real number except $$x = 1$$.

At any point $$c \lt 1$$: $$f(c) = c + 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(x + 2) = c + 2 = f(c)$$. So $$f$$ is continuous at every $$c \lt 1$$.

At any point $$c \gt 1$$: $$f(c) = c - 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(x - 2) = c - 2 = f(c)$$. So $$f$$ is continuous at every $$c \gt 1$$.

At $$c = 1$$: By the middle line of the definition (the case $$x = 1$$), $$f(1) = 0$$.

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x + 2) = 3$$.

Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+}(x - 2) = -1$$.

The left and right hand limits are unequal, so $$\lim_{x \to 1} f(x)$$ does not exist. Hence $$\lim_{x \to 1} f(x) \ne f(1)$$ and $$f$$ is not continuous at $$x = 1$$.

Hence the only point of discontinuity is $$x = 1$$.

The function $$f$$ is defined for all $$x \ne 0$$; its domain is $$\mathbf{R} - \{0\}$$.

At any point $$c \lt 0$$: Here $$f(c) = c + 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(x + 2) = c + 2 = f(c)$$. So $$f$$ is continuous at every $$c \lt 0$$.

At any point $$c \gt 0$$: Here $$f(c) = -c + 2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c}(-x + 2) = -c + 2 = f(c)$$. So $$f$$ is continuous at every $$c \gt 0$$.

Thus $$f$$ is continuous at every point of its domain. Since $$x = 0$$ is not in the domain of $$f$$, continuity is not discussed there.

Hence $$f$$ is a continuous function; it has no point of discontinuity in its domain.

At any point $$c \gt 0$$: $$f(c) = c$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c} x = c = f(c)$$. So $$f$$ is continuous at every $$c \gt 0$$.

At any point $$c \lt 0$$: $$f(c) = c^2$$ and $$\lim_{x \to c} f(x) = \lim_{x \to c} x^2 = c^2 = f(c)$$. So $$f$$ is continuous at every $$c \lt 0$$.

At $$c = 0$$: $$f(0) = 0$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0$$.

Right hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$$.

So $$\lim_{x \to 0} f(x) = 0 = f(0)$$, and $$f$$ is continuous at $$x = 0$$.

Hence $$f$$ is continuous at every real number; it is a continuous function.

Recall that a function $$p$$ is a polynomial function if it has the form $$p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$ for some natural number $$n$$, where $$a_0, a_1, \ldots, a_n$$ are real constants.

Let $$c$$ be any real number. By the algebra of limits, $$\lim_{x \to c} x^k = c^k$$ for each $$k$$, and the limit of a sum is the sum of the limits. Hence

$$\lim_{x \to c} p(x) = a_0 + a_1 c + a_2 c^2 + \cdots + a_n c^n = p(c)$$

Thus $$\lim_{x \to c} p(x) = p(c)$$ holds for every real number $$c$$.

Therefore every polynomial function is continuous at every real number, i.e. it is a continuous function.

The greatest integer function $$f(x) = [x]$$ assigns to each $$x$$ the greatest integer not exceeding $$x$$.

At a non-integer point $$c$$: there is an integer $$n$$ with $$n \lt c \lt n + 1$$. For every $$x$$ in the open interval $$(n,\ n + 1)$$, that is for $$n \lt x \lt n + 1$$, we have $$[x] = n$$. Hence $$\lim_{x \to c} f(x) = n = [c] = f(c)$$, so $$f$$ is continuous at every non-integer point.

At an integer $$c = n$$: $$f(n) = [n] = n$$.

Left hand limit: for $$x$$ slightly less than $$n$$, that is for $$n - 1 \lt x \lt n$$, we have $$[x] = n - 1$$, so $$\lim_{x \to n^-} f(x) = n - 1$$.

Right hand limit: for $$x$$ slightly more than $$n$$, that is for $$n \lt x \lt n + 1$$, we have $$[x] = n$$, so $$\lim_{x \to n^+} f(x) = n$$.

Since $$n - 1 \ne n$$, $$\lim_{x \to n} f(x)$$ does not exist, so $$f$$ is not continuous at $$x = n$$.

Hence $$f(x) = [x]$$ is discontinuous at every integer and continuous everywhere else.

A rational function is a function of the form $$f(x) = \dfrac{p(x)}{q(x)}$$, where $$p$$ and $$q$$ are polynomial functions and $$q$$ is not the zero polynomial. The domain of $$f$$ is the set of all real numbers $$x$$ for which $$q(x) \ne 0$$.

Let $$c$$ be any point of the domain, so $$q(c) \ne 0$$. Since polynomials are continuous (Example 14), we have $$\lim_{x \to c} p(x) = p(c)$$ and $$\lim_{x \to c} q(x) = q(c)$$.

By the quotient rule for limits, since $$\lim_{x \to c} q(x) = q(c) \ne 0$$,

$$\lim_{x \to c} f(x) = \lim_{x \to c} \dfrac{p(x)}{q(x)} = \dfrac{p(c)}{q(c)} = f(c)$$

Hence $$f$$ is continuous at every point $$c$$ of its domain. Therefore every rational function is continuous.

Let $$f(x) = \sin x$$ and let $$c$$ be any real number. Put $$x = c + h$$, so that $$x \to c$$ is equivalent to $$h \to 0$$. Then

$$\lim_{x \to c} \sin x = \lim_{h \to 0} \sin(c + h)$$

$$= \lim_{h \to 0} [\sin c \cos h + \cos c \sin h]$$

$$= \sin c \left(\lim_{h \to 0} \cos h\right) + \cos c \left(\lim_{h \to 0} \sin h\right)$$

Since $$\lim_{h \to 0} \cos h = 1$$ and $$\lim_{h \to 0} \sin h = 0$$,

$$\lim_{x \to c} \sin x = \sin c \cdot 1 + \cos c \cdot 0 = \sin c = f(c)$$

Thus $$\lim_{x \to c} f(x) = f(c)$$ for every real $$c$$, so the sine function is continuous at every real number.

We have $$\tan x = \dfrac{\sin x}{\cos x}$$, which is defined for all $$x$$ with $$\cos x \ne 0$$, i.e. for $$x \ne (2n + 1)\dfrac{\pi}{2}$$, $$n \in \mathbf{Z}$$. This set is the domain of $$f$$.

The sine function and the cosine function are each continuous at every real number.

By the quotient rule for continuous functions, the quotient $$\dfrac{\sin x}{\cos x}$$ is continuous at every point where the denominator $$\cos x \ne 0$$.

Hence $$f(x) = \tan x$$ is continuous at every point of its domain, i.e. it is a continuous function.

Write $$f(x) = \sin(x^2)$$ as the composition $$f = g \circ h$$, where $$h(x) = x^2$$ and $$g(t) = \sin t$$, so that $$f(x) = g(h(x)) = \sin(x^2)$$.

The function $$h(x) = x^2$$ is a polynomial, hence continuous everywhere.

The function $$g(t) = \sin t$$ is continuous everywhere.

By the theorem that the composition of two continuous functions is continuous, $$f = g \circ h$$ is continuous everywhere.

Hence $$f(x) = \sin(x^2)$$ is a continuous function.

Define $$g(x) = 1 - x + |x|$$ and $$h(t) = |t|$$, so that $$f(x) = h(g(x)) = |1 - x + |x||$$.

$$g$$ is continuous: The function $$1 - x$$ is a polynomial, hence continuous, and $$|x|$$ is continuous (Example 7). The sum of two continuous functions is continuous, so $$g(x) = (1 - x) + |x|$$ is continuous everywhere.

$$h$$ is continuous: $$h(t) = |t|$$, the modulus function, is continuous everywhere (Example 7).

Since $$f = h \circ g$$ is the composition of two continuous functions, $$f$$ is continuous everywhere.

Hence $$f(x) = |1 - x + |x||$$ is a continuous function.

The function $$f(x) = 5x - 3$$ is a polynomial, so for any real $$c$$, $$\lim_{x \to c} f(x) = 5c - 3$$.

At $$x = 0$$: $$f(0) = 5(0) - 3 = -3$$ and $$\lim_{x \to 0}(5x - 3) = -3$$. Since they are equal, $$f$$ is continuous at $$x = 0$$.

At $$x = -3$$: $$f(-3) = 5(-3) - 3 = -18$$ and $$\lim_{x \to -3}(5x - 3) = -18$$. Since they are equal, $$f$$ is continuous at $$x = -3$$.

At $$x = 5$$: $$f(5) = 5(5) - 3 = 22$$ and $$\lim_{x \to 5}(5x - 3) = 22$$. Since they are equal, $$f$$ is continuous at $$x = 5$$.

Hence $$f$$ is continuous at each of the three points.

$$f(x) = 2x^2 - 1$$ is a polynomial function.

Value at $$x = 3$$: $$f(3) = 2(3)^2 - 1 = 18 - 1 = 17$$.

Limit as $$x \to 3$$: $$\lim_{x \to 3} f(x) = \lim_{x \to 3}(2x^2 - 1) = 2(3)^2 - 1 = 17$$.

Since $$\lim_{x \to 3} f(x) = 17 = f(3)$$, the function $$f$$ is continuous at $$x = 3$$.

$$f(x) = x^n$$, where $$n$$ is a positive integer, is a polynomial function.

Value at $$x = n$$: $$f(n) = n^n$$.

Limit as $$x \to n$$: $$\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$$.

Since $$\lim_{x \to n} f(x) = n^n = f(n)$$, the function $$f$$ is continuous at $$x = n$$.

At $$x = 0$$: Since $$0 \le 1$$, $$f(0) = 0$$. For $$x$$ near $$0$$ we have $$x \le 1$$, so $$\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0 = f(0)$$. Hence $$f$$ is continuous at $$x = 0$$.

At $$x = 1$$: $$f(1) = 1$$. Left hand limit $$= \lim_{x \to 1^-} x = 1$$; right hand limit $$= \lim_{x \to 1^+} 5 = 5$$. Since $$1 \ne 5$$, $$\lim_{x \to 1} f(x)$$ does not exist, so $$f$$ is not continuous at $$x = 1$$.

At $$x = 2$$: Since $$2 > 1$$, $$f(2) = 5$$. For $$x$$ near $$2$$ we have $$x > 1$$, so $$\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5 = f(2)$$. Hence $$f$$ is continuous at $$x = 2$$.

For $$x \lt 2$$ and for $$x \gt 2$$, the function $$f$$ is given by polynomial expressions, hence it is continuous at every such point. Only the junction $$x = 2$$ needs checking.

At $$x = 2$$: Since $$2 \le 2$$, $$f(2) = 2(2) + 3 = 7$$.

Left hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-}(2x + 3) = 7$$.

Right hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+}(2x - 3) = 2(2) - 3 = 1$$.

Since the left and right hand limits are unequal, $$\lim_{x \to 2} f(x)$$ does not exist, so $$f$$ is discontinuous at $$x = 2$$.

Hence $$x = 2$$ is the only point of discontinuity.

On each of the regions $$x \lt -3$$, $$-3 \lt x \lt 3$$ and $$x \gt 3$$, the function is given by a continuous expression (note that $$|x| + 3 = -x + 3$$ for $$x \lt -3$$). So $$f$$ is continuous at every point except possibly the junctions $$x = -3$$ and $$x = 3$$.

At $$x = -3$$: $$f(-3) = |-3| + 3 = 6$$.

Left hand limit: $$\lim_{x \to -3^-}(|x| + 3) = \lim_{x \to -3^-}(-x + 3) = 3 + 3 = 6$$.

Right hand limit: $$\lim_{x \to -3^+}(-2x) = -2(-3) = 6$$.

All three values equal $$6$$, so $$f$$ is continuous at $$x = -3$$.

At $$x = 3$$: $$f(3) = 6(3) + 2 = 20$$.

Left hand limit: $$\lim_{x \to 3^-}(-2x) = -2(3) = -6$$.

Right hand limit: $$\lim_{x \to 3^+}(6x + 2) = 20$$.

Since $$-6 \ne 20$$, $$f$$ is discontinuous at $$x = 3$$.

Hence the only point of discontinuity is $$x = 3$$.

For $$x \gt 0$$, $$|x| = x$$, so $$f(x) = \dfrac{x}{x} = 1$$. For $$x \lt 0$$, $$|x| = -x$$, so $$f(x) = \dfrac{-x}{x} = -1$$. By definition $$f(0) = 0$$.

At any $$c \gt 0$$: $$f$$ is constantly $$1$$ near $$c$$, so it is continuous there.

At any $$c \lt 0$$: $$f$$ is constantly $$-1$$ near $$c$$, so it is continuous there.

At $$x = 0$$: $$f(0) = 0$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = -1$$.

Right hand limit: $$\lim_{x \to 0^+} f(x) = 1$$.

Since $$-1 \ne 1$$, $$\lim_{x \to 0} f(x)$$ does not exist, so $$f$$ is discontinuous at $$x = 0$$.

Hence the only point of discontinuity is $$x = 0$$.

For $$x < 0$$, $$|x| = -x$$, so $$f(x) = \dfrac{x}{|x|} = \dfrac{x}{-x} = -1$$. For $$x \ge 0$$, $$f(x) = -1$$ by definition.

Therefore $$f(x) = -1$$ for every real number $$x$$, i.e. $$f$$ is a constant function.

A constant function is continuous at every real number. In particular, at the junction $$x = 0$$: $$f(0) = -1$$, $$\lim_{x \to 0^-} f(x) = -1$$ and $$\lim_{x \to 0^+} f(x) = -1$$, so $$f$$ is continuous at $$x = 0$$ as well.

Hence $$f$$ has no point of discontinuity.

For $$x \lt 1$$, $$f(x) = x^2 + 1$$, and for $$x \gt 1$$, $$f(x) = x + 1$$; both are polynomial expressions, so $$f$$ is continuous at every point other than possibly $$x = 1$$.

At $$x = 1$$: Since $$1 \ge 1$$, $$f(1) = 1 + 1 = 2$$.

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x^2 + 1) = 1 + 1 = 2$$.

Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+}(x + 1) = 1 + 1 = 2$$.

So $$\lim_{x \to 1} f(x) = 2 = f(1)$$, and $$f$$ is continuous at $$x = 1$$.

Hence $$f$$ is continuous everywhere; there is no point of discontinuity.

For $$x \lt 2$$ and $$x \gt 2$$, $$f$$ is given by polynomial expressions, so it is continuous there. Only $$x = 2$$ needs checking.

At $$x = 2$$: Since $$2 \le 2$$, $$f(2) = 2^3 - 3 = 5$$.

Left hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-}(x^3 - 3) = 8 - 3 = 5$$.

Right hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+}(x^2 + 1) = 4 + 1 = 5$$.

So $$\lim_{x \to 2} f(x) = 5 = f(2)$$, and $$f$$ is continuous at $$x = 2$$.

Hence $$f$$ is continuous everywhere; there is no point of discontinuity.

For $$x \lt 1$$ and $$x \gt 1$$, $$f$$ is given by polynomial expressions, so it is continuous there. Only $$x = 1$$ needs checking.

At $$x = 1$$: Since $$1 \le 1$$, $$f(1) = 1^{10} - 1 = 0$$.

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x^{10} - 1) = 1 - 1 = 0$$.

Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1$$.

Since $$0 \ne 1$$, the left and right hand limits are unequal, so $$\lim_{x \to 1} f(x)$$ does not exist and $$f$$ is discontinuous at $$x = 1$$.

Hence the only point of discontinuity is $$x = 1$$.

For $$x \lt 1$$ and $$x \gt 1$$, $$f$$ is given by polynomial expressions, hence continuous. We check the junction $$x = 1$$.

At $$x = 1$$: Since $$1 \le 1$$, $$f(1) = 1 + 5 = 6$$.

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x + 5) = 6$$.

Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+}(x - 5) = 1 - 5 = -4$$.

Since $$6 \ne -4$$, $$\lim_{x \to 1} f(x)$$ does not exist, so $$f$$ is discontinuous at $$x = 1$$.

Since $$f$$ fails to be continuous at $$x = 1$$, it is not a continuous function.

The domain of $$f$$ is $$[0, 10]$$. On the intervals $$0 \le x \le 1$$, $$1 \lt x \lt 3$$ and $$3 \le x \le 10$$ the function is constant, hence continuous at every interior point. We examine the junctions $$x = 1$$ and $$x = 3$$.

At $$x = 1$$: $$f(1) = 3$$ (since $$0 \le 1 \le 1$$).

Left hand limit: $$\lim_{x \to 1^-} f(x) = 3$$. Right hand limit: $$\lim_{x \to 1^+} f(x) = 4$$.

Since $$3 \ne 4$$, $$f$$ is discontinuous at $$x = 1$$.

At $$x = 3$$: $$f(3) = 5$$ (since $$3 \le 3 \le 10$$).

Left hand limit: $$\lim_{x \to 3^-} f(x) = 4$$. Right hand limit: $$\lim_{x \to 3^+} f(x) = 5$$.

Since $$4 \ne 5$$, $$f$$ is discontinuous at $$x = 3$$.

Hence $$f$$ is continuous on $$[0, 10]$$ except at the two points $$x = 1$$ and $$x = 3$$.

On the regions $$x \lt 0$$, $$0 \le x \le 1$$ and $$x \gt 1$$, $$f$$ is given by continuous expressions, so it is continuous there. We examine the junctions $$x = 0$$ and $$x = 1$$.

At $$x = 0$$: $$f(0) = 0$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x = 0$$. Right hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$$.

So $$\lim_{x \to 0} f(x) = 0 = f(0)$$; $$f$$ is continuous at $$x = 0$$.

At $$x = 1$$: $$f(1) = 0$$ (since $$0 \le 1 \le 1$$).

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0 = 0$$. Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4x = 4$$.

Since $$0 \ne 4$$, $$f$$ is discontinuous at $$x = 1$$.

Hence $$f$$ is continuous everywhere except at $$x = 1$$.

On the regions $$x \lt -1$$, $$-1 \lt x \le 1$$ and $$x \gt 1$$, $$f$$ is given by continuous expressions, so it is continuous there. We examine the junctions $$x = -1$$ and $$x = 1$$.

At $$x = -1$$: $$f(-1) = -2$$ (since $$-1 \le -1$$).

Left hand limit: $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-}(-2) = -2$$. Right hand limit: $$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} 2x = -2$$.

So $$\lim_{x \to -1} f(x) = -2 = f(-1)$$; $$f$$ is continuous at $$x = -1$$.

At $$x = 1$$: $$f(1) = 2(1) = 2$$ (since $$-1 \lt 1 \le 1$$).

Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2x = 2$$. Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2 = 2$$.

So $$\lim_{x \to 1} f(x) = 2 = f(1)$$; $$f$$ is continuous at $$x = 1$$.

Hence $$f$$ is continuous at every real number.

For $$f$$ to be continuous at $$x = 3$$, the left hand limit, the right hand limit and $$f(3)$$ must all be equal.

Value: Since $$3 \le 3$$, $$f(3) = a(3) + 1 = 3a + 1$$.

Left hand limit: $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-}(ax + 1) = 3a + 1$$.

Right hand limit: $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+}(bx + 3) = 3b + 3$$.

Continuity at $$x = 3$$ requires $$3a + 1 = 3b + 3$$, that is

$$3a = 3b + 2 \quad\Rightarrow\quad a = b + \dfrac{2}{3}$$

This is the required relationship between $$a$$ and $$b$$.

Continuity at $$x = 0$$:

Value: $$f(0) = \lambda(0^2 - 2 \cdot 0) = 0$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0 - 0) = 0$$.

Right hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+}(4x + 1) = 1$$.

For continuity at $$x = 0$$ the right hand limit must equal $$f(0)$$, i.e. $$1 = 0$$, which is impossible. Hence there is no value of $$\lambda$$ for which $$f$$ is continuous at $$x = 0$$.

Continuity at $$x = 1$$:

The point $$x = 1$$ lies in the region $$x > 0$$, where $$f(x) = 4x + 1$$, a polynomial. Hence $$\lim_{x \to 1} f(x) = 4(1) + 1 = 5 = f(1)$$, so $$f$$ is continuous at $$x = 1$$ for every value of $$\lambda$$.

Let $$n$$ be any integer. By definition $$g(x) = x - [x]$$, so $$g(n) = n - [n] = n - n = 0$$.

Left hand limit at $$x = n$$: For $$x$$ slightly less than $$n$$, that is for $$n - 1 \lt x \lt n$$, the greatest integer not exceeding $$x$$ is $$[x] = n - 1$$. Hence

$$\lim_{x \to n^-} g(x) = \lim_{x \to n^-}(x - [x]) = \lim_{x \to n^-}(x - (n - 1)) = n - (n - 1) = 1$$

Right hand limit at $$x = n$$: For $$x$$ slightly more than $$n$$, that is for $$n \lt x \lt n + 1$$, we have $$[x] = n$$. Hence

$$\lim_{x \to n^+} g(x) = \lim_{x \to n^+}(x - n) = n - n = 0$$

Since the left hand limit ($$1$$) and the right hand limit ($$0$$) are unequal, $$\lim_{x \to n} g(x)$$ does not exist.

Therefore $$g$$ is discontinuous at $$x = n$$. As $$n$$ was an arbitrary integer, $$g(x) = x - [x]$$ is discontinuous at all integral points.

The function $$f(x) = x^2 - \sin x + 5$$ is built from $$x^2$$ (a polynomial), $$\sin x$$, and the constant $$5$$.

Each of $$x^2$$, $$\sin x$$ and the constant function is continuous at every real number, in particular at $$x = \pi$$. A difference and a sum of continuous functions is continuous, so $$f$$ is continuous at $$x = \pi$$.

To confirm directly: $$f(\pi) = \pi^2 - \sin \pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5$$, and

$$\lim_{x \to \pi} f(x) = \pi^2 - \sin \pi + 5 = \pi^2 + 5 = f(\pi)$$

Hence $$f$$ is continuous at $$x = \pi$$.

It is known that $$\sin x$$ is continuous at every real number.

Cosine function: For any real $$c$$, putting $$x = c + h$$, $$\cos(c + h) = \cos c \cos h - \sin c \sin h \to \cos c \cdot 1 - \sin c \cdot 0 = \cos c$$ as $$h \to 0$$. Hence $$\cos x$$ is continuous at every real number.

Cosecant function: $$\operatorname{cosec} x = \dfrac{1}{\sin x}$$. As a quotient of continuous functions it is continuous wherever the denominator $$\sin x \ne 0$$, i.e. for $$x \ne n\pi$$, $$n \in \mathbf{Z}$$. (At $$x = n\pi$$ it is not defined.)

Secant function: $$\sec x = \dfrac{1}{\cos x}$$ is continuous wherever $$\cos x \ne 0$$, i.e. for $$x \ne (2n + 1)\dfrac{\pi}{2}$$, $$n \in \mathbf{Z}$$.

Cotangent function: $$\cot x = \dfrac{\cos x}{\sin x}$$ is continuous wherever $$\sin x \ne 0$$, i.e. for $$x \ne n\pi$$, $$n \in \mathbf{Z}$$.

Thus each of these four functions is continuous at every point of its domain.

For $$x \lt 0$$, $$f(x) = \dfrac{\sin x}{x}$$ is a quotient of continuous functions with denominator $$x \ne 0$$, hence continuous. For $$x \gt 0$$, $$f(x) = x + 1$$ is a polynomial, hence continuous. Only the junction $$x = 0$$ needs checking.

At $$x = 0$$: Since $$0 \ge 0$$, $$f(0) = 0 + 1 = 1$$.

Right hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+}(x + 1) = 1$$.

Left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \dfrac{\sin x}{x} = 1$$, using the standard limit $$\lim_{x \to 0} \dfrac{\sin x}{x} = 1$$.

So $$\lim_{x \to 0} f(x) = 1 = f(0)$$, and $$f$$ is continuous at $$x = 0$$.

Hence $$f$$ is continuous everywhere; there is no point of discontinuity.

For $$x \ne 0$$, $$f(x) = x^2 \sin \dfrac{1}{x}$$ is a product of $$x^2$$ (a polynomial) and $$\sin \dfrac{1}{x}$$ (a composition of continuous functions, defined for $$x \ne 0$$), hence $$f$$ is continuous at every $$x \ne 0$$.

At $$x = 0$$: $$f(0) = 0$$.

For all $$x \ne 0$$, since $$\left|\sin \dfrac{1}{x}\right| \le 1$$,

$$0 \le \left|x^2 \sin \dfrac{1}{x}\right| \le x^2$$

As $$x \to 0$$, $$x^2 \to 0$$, so by the Sandwich theorem $$\lim_{x \to 0} x^2 \sin \dfrac{1}{x} = 0$$.

Thus $$\lim_{x \to 0} f(x) = 0 = f(0)$$, so $$f$$ is continuous at $$x = 0$$.

Hence $$f$$ is continuous at every real number; it is a continuous function.

For $$x \ne 0$$, $$f(x) = \sin x - \cos x$$ is a difference of continuous functions, hence continuous at every $$x \ne 0$$.

At $$x = 0$$: By definition $$f(0) = -1$$.

$$\lim_{x \to 0} f(x) = \lim_{x \to 0}(\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$$

So $$\lim_{x \to 0} f(x) = -1 = f(0)$$, and $$f$$ is continuous at $$x = 0$$.

Hence $$f$$ is continuous at every real number; it is a continuous function.

For continuity at $$x = \dfrac{\pi}{2}$$, we need $$\lim_{x \to \pi/2} f(x) = f\left(\dfrac{\pi}{2}\right) = 3$$.

Compute the limit. Put $$x = \dfrac{\pi}{2} + h$$, so $$h \to 0$$ as $$x \to \dfrac{\pi}{2}$$. Then

$$\cos x = \cos\left(\dfrac{\pi}{2} + h\right) = -\sin h, \qquad \pi - 2x = \pi - 2\left(\dfrac{\pi}{2} + h\right) = -2h$$

$$\lim_{x \to \pi/2} \dfrac{k \cos x}{\pi - 2x} = \lim_{h \to 0} \dfrac{k(-\sin h)}{-2h} = \dfrac{k}{2} \lim_{h \to 0} \dfrac{\sin h}{h} = \dfrac{k}{2} \cdot 1 = \dfrac{k}{2}$$

Setting this equal to $$3$$: $$\dfrac{k}{2} = 3$$, hence $$k = 6$$.

For $$x \le 2$$, $$f(x) = kx^2$$; for $$x > 2$$, $$f(x) = 3$$. For continuity at $$x = 2$$, the left hand limit, the right hand limit and $$f(2)$$ must all be equal.

Value: $$f(2) = k(2)^2 = 4k$$.

Left hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} kx^2 = 4k$$.

Right hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 3 = 3$$.

Continuity requires $$4k = 3$$, hence $$k = \dfrac{3}{4}$$.

For continuity at $$x = \pi$$, the left hand limit, the right hand limit and $$f(\pi)$$ must all be equal.

Value: $$f(\pi) = k\pi + 1$$ (since $$\pi \le \pi$$).

Left hand limit: $$\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-}(kx + 1) = k\pi + 1$$.

Right hand limit: $$\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos \pi = -1$$.

Continuity requires $$k\pi + 1 = -1$$, so $$k\pi = -2$$, giving $$k = -\dfrac{2}{\pi}$$.

For continuity at $$x = 5$$, the left hand limit, the right hand limit and $$f(5)$$ must all be equal.

Value: $$f(5) = k(5) + 1 = 5k + 1$$ (since $$5 \le 5$$).

Left hand limit: $$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-}(kx + 1) = 5k + 1$$.

Right hand limit: $$\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+}(3x - 5) = 3(5) - 5 = 10$$.

Continuity requires $$5k + 1 = 10$$, so $$5k = 9$$, giving $$k = \dfrac{9}{5}$$.

On the three regions $$x \le 2$$, $$2 \lt x \lt 10$$ and $$x \ge 10$$, $$f$$ is given by continuous expressions. For $$f$$ to be continuous everywhere it must be continuous at the junctions $$x = 2$$ and $$x = 10$$.

Continuity at $$x = 2$$: $$f(2) = 5$$.

Left hand limit $$= \lim_{x \to 2^-} 5 = 5$$; right hand limit $$= \lim_{x \to 2^+}(ax + b) = 2a + b$$.

Continuity gives $$2a + b = 5$$.  ...(1)

Continuity at $$x = 10$$: $$f(10) = 21$$.

Left hand limit $$= \lim_{x \to 10^-}(ax + b) = 10a + b$$; right hand limit $$= \lim_{x \to 10^+} 21 = 21$$.

Continuity gives $$10a + b = 21$$.  ...(2)

Subtracting (1) from (2): $$8a = 16$$, so $$a = 2$$. Substituting into (1): $$2(2) + b = 5$$, so $$b = 1$$.

Hence $$a = 2$$ and $$b = 1$$.

Write $$f(x) = \cos(x^2)$$ as the composition $$f = g \circ h$$, where $$h(x) = x^2$$ and $$g(t) = \cos t$$.

$$h(x) = x^2$$ is a polynomial, hence continuous everywhere; $$g(t) = \cos t$$ is continuous everywhere.

Since the composition of two continuous functions is continuous, $$f = g \circ h$$ is continuous everywhere.

Hence $$f(x) = \cos(x^2)$$ is a continuous function.

Write $$f(x) = |\cos x|$$ as the composition $$f = g \circ h$$, where $$h(x) = \cos x$$ and $$g(t) = |t|$$.

$$h(x) = \cos x$$ is continuous everywhere; $$g(t) = |t|$$, the modulus function, is continuous everywhere.

Since the composition of two continuous functions is continuous, $$f = g \circ h$$ is continuous everywhere.

Hence $$f(x) = |\cos x|$$ is a continuous function.

Write $$f(x) = \sin |x|$$ as the composition $$f = g \circ h$$, where $$h(x) = |x|$$ and $$g(t) = \sin t$$.

$$h(x) = |x|$$, the modulus function, is continuous everywhere; $$g(t) = \sin t$$ is continuous everywhere.

Since the composition of two continuous functions is continuous, $$f = g \circ h$$ is continuous everywhere.

Hence $$f(x) = \sin |x|$$ is a continuous function.

Both $$|x|$$ and $$|x + 1|$$ are modulus functions, which are continuous at every real number. The difference of two continuous functions is continuous, so $$f(x) = |x| - |x + 1|$$ is continuous everywhere.

To see this in detail, note $$|x|$$ changes form at $$x = 0$$ and $$|x + 1|$$ at $$x = -1$$. Splitting into cases:

• For $$x \lt -1$$: $$|x| = -x$$ and $$|x + 1| = -(x + 1)$$, so $$f(x) = -x + (x + 1) = 1$$.
• For $$-1 \le x \lt 0$$: $$|x| = -x$$ and $$|x + 1| = x + 1$$, so $$f(x) = -x - (x + 1) = -2x - 1$$.
• For $$x \ge 0$$: $$|x| = x$$, $$|x + 1| = x + 1$$, so $$f(x) = x - (x + 1) = -1$$.

At $$x = -1$$: the value from the left is $$1$$, and $$-2x - 1$$ at $$x = -1$$ also gives $$1$$; further $$f(-1) = 1$$. Continuous.

At $$x = 0$$: $$-2x - 1$$ at $$x = 0$$ gives $$-1$$, matching the right piece $$-1$$; further $$f(0) = -1$$. Continuous.

Each piece is a polynomial and the pieces agree at the joins, so $$f$$ is continuous everywhere.

Hence $$f$$ has no point of discontinuity.

$$f(x) = \sin(x^2)$$ is the composition of the inner function $$u = x^2$$ and the outer function $$\sin u$$. Apply the chain rule.

$$f'(x) = \dfrac{d}{dx}\sin(x^2) = \cos(x^2) \cdot \dfrac{d}{dx}(x^2)$$

$$= \cos(x^2) \cdot 2x = 2x \cos(x^2)$$

Let $$y = \sin(x^2 + 5)$$. By the chain rule, with inner function $$u = x^2 + 5$$,

$$\dfrac{dy}{dx} = \cos(x^2 + 5) \cdot \dfrac{d}{dx}(x^2 + 5) = \cos(x^2 + 5) \cdot 2x$$

$$\dfrac{dy}{dx} = 2x \cos(x^2 + 5)$$

Let $$y = \cos(\sin x)$$. By the chain rule, with inner function $$u = \sin x$$,

$$\dfrac{dy}{dx} = -\sin(\sin x) \cdot \dfrac{d}{dx}(\sin x) = -\sin(\sin x) \cdot \cos x$$

$$\dfrac{dy}{dx} = -\cos x \, \sin(\sin x)$$

Let $$y = \sin(ax + b)$$. By the chain rule, with inner function $$u = ax + b$$,

$$\dfrac{dy}{dx} = \cos(ax + b) \cdot \dfrac{d}{dx}(ax + b) = \cos(ax + b) \cdot a$$

$$\dfrac{dy}{dx} = a \cos(ax + b)$$

Let $$y = \sec(\tan \sqrt{x})$$. This is a composition of three functions; differentiate from the outside inward.

$$\dfrac{dy}{dx} = \sec(\tan \sqrt{x})\,\tan(\tan \sqrt{x}) \cdot \dfrac{d}{dx}(\tan \sqrt{x})$$

Now $$\dfrac{d}{dx}(\tan \sqrt{x}) = \sec^2 \sqrt{x} \cdot \dfrac{d}{dx}\sqrt{x} = \sec^2 \sqrt{x} \cdot \dfrac{1}{2\sqrt{x}}$$.

Therefore

$$\dfrac{dy}{dx} = \dfrac{\sec^2 \sqrt{x}\,\sec(\tan \sqrt{x})\,\tan(\tan \sqrt{x})}{2\sqrt{x}}$$

Let $$y = \dfrac{\sin(ax + b)}{\cos(cx + d)}$$. Write $$u = \sin(ax + b)$$ and $$v = \cos(cx + d)$$.

By the chain rule, $$u' = a\cos(ax + b)$$ and $$v' = -c\sin(cx + d)$$.

By the quotient rule, $$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2}$$:

$$\dfrac{dy}{dx} = \dfrac{a\cos(ax + b)\cos(cx + d) - \sin(ax + b)\,[-c\sin(cx + d)]}{\cos^2(cx + d)}$$

$$\dfrac{dy}{dx} = \dfrac{a\cos(ax + b)\cos(cx + d) + c\sin(ax + b)\sin(cx + d)}{\cos^2(cx + d)}$$

Let $$y = \cos x^3 \cdot \sin^2(x^5)$$, the product of $$u = \cos(x^3)$$ and $$v = \sin^2(x^5)$$.

By the chain rule,

$$u' = \dfrac{d}{dx}\cos(x^3) = -\sin(x^3) \cdot 3x^2 = -3x^2 \sin(x^3)$$

$$v' = \dfrac{d}{dx}[\sin(x^5)]^2 = 2\sin(x^5) \cdot \cos(x^5) \cdot 5x^4 = 10x^4 \sin(x^5)\cos(x^5)$$

By the product rule, $$\dfrac{dy}{dx} = u'v + uv'$$:

$$\dfrac{dy}{dx} = -3x^2 \sin(x^3)\sin^2(x^5) + 10x^4 \sin(x^5)\cos(x^5)\cos(x^3)$$

Let $$y = 2\sqrt{\cot(x^2)} = 2[\cot(x^2)]^{1/2}$$.

By the chain rule,

$$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{2}[\cot(x^2)]^{-1/2} \cdot \dfrac{d}{dx}\cot(x^2)$$

Now $$\dfrac{d}{dx}\cot(x^2) = -\operatorname{cosec}^2(x^2) \cdot 2x$$. Therefore

$$\dfrac{dy}{dx} = [\cot(x^2)]^{-1/2} \cdot \left(-2x\operatorname{cosec}^2(x^2)\right) = \dfrac{-2x\operatorname{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$$

Let $$y = \cos \sqrt{x}$$. By the chain rule, with inner function $$u = \sqrt{x}$$,

$$\dfrac{dy}{dx} = -\sin \sqrt{x} \cdot \dfrac{d}{dx}\sqrt{x} = -\sin \sqrt{x} \cdot \dfrac{1}{2\sqrt{x}}$$

$$\dfrac{dy}{dx} = \dfrac{-\sin \sqrt{x}}{2\sqrt{x}}$$

The function is $$f(x) = |x - 1| = \begin{cases} x - 1, & \text{if } x \ge 1 \\ 1 - x, & \text{if } x \lt 1 \end{cases}$$, and $$f(1) = 0$$.

The derivative at $$x = 1$$ exists only if the left hand derivative equals the right hand derivative.

Left hand derivative: for $$h \lt 0$$ the point $$1 + h$$ satisfies $$1 + h \lt 1$$, so $$f(1 + h) = |1 + h - 1| = |h| = -h$$. Hence

$$\lim_{h \to 0^-} \dfrac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \dfrac{-h - 0}{h} = \lim_{h \to 0^-}(-1) = -1$$

Right hand derivative: for $$h \gt 0$$ the point $$1 + h$$ satisfies $$1 + h \gt 1$$, so $$f(1 + h) = |1 + h - 1| = |h| = h$$. Hence

$$\lim_{h \to 0^+} \dfrac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \dfrac{h - 0}{h} = \lim_{h \to 0^+}(1) = 1$$

Since the left hand derivative $$(-1)$$ and the right hand derivative $$(1)$$ are not equal, $$f'(1)$$ does not exist.

Hence $$f(x) = |x - 1|$$ is not differentiable at $$x = 1$$.

A function differentiable at a point must be continuous there. The greatest integer function is discontinuous at every integer, so it cannot be differentiable at $$x = 1$$ or $$x = 2$$. We confirm this through the derivative quotient.

At $$x = 1$$: $$f(1) = [1] = 1$$.

Right hand derivative: for small $$h \gt 0$$, the point $$1 + h$$ lies in $$(1,\ 2)$$, so $$[1 + h] = 1$$. Then $$\dfrac{f(1 + h) - f(1)}{h} = \dfrac{1 - 1}{h} = 0$$, so the right hand derivative is $$0$$.

Left hand derivative: for small $$h \lt 0$$, the point $$1 + h$$ lies in $$(0,\ 1)$$, so $$[1 + h] = 0$$. Then $$\dfrac{f(1 + h) - f(1)}{h} = \dfrac{0 - 1}{h} = \dfrac{-1}{h}$$, and as $$h \to 0^-$$ this tends to $$+\infty$$. So the left hand derivative does not exist as a finite number.

Since the left hand derivative does not exist finitely while the right hand derivative is $$0$$, $$f$$ is not differentiable at $$x = 1$$.

At $$x = 2$$: $$f(2) = [2] = 2$$.

Right hand derivative: for small $$h \gt 0$$, the point $$2 + h$$ lies in $$(2,\ 3)$$, so $$[2 + h] = 2$$. Then $$\dfrac{f(2 + h) - f(2)}{h} = \dfrac{2 - 2}{h} = 0$$, so the right hand derivative is $$0$$.

Left hand derivative: for small $$h \lt 0$$, the point $$2 + h$$ lies in $$(1,\ 2)$$, so $$[2 + h] = 1$$. Then $$\dfrac{f(2 + h) - f(2)}{h} = \dfrac{1 - 2}{h} = \dfrac{-1}{h} \to +\infty$$ as $$h \to 0^-$$; again the left hand derivative does not exist finitely.

Hence $$f$$ is not differentiable at $$x = 2$$ either.

Given the relation $$x - y = \pi$$. Differentiate both sides with respect to $$x$$, treating $$y$$ as a function of $$x$$.

$$\dfrac{d}{dx}(x) - \dfrac{d}{dx}(y) = \dfrac{d}{dx}(\pi)$$

$$1 - \dfrac{dy}{dx} = 0$$

$$\dfrac{dy}{dx} = 1$$

Given $$y + \sin y = \cos x$$. Differentiate both sides with respect to $$x$$:

$$\dfrac{dy}{dx} + \cos y \cdot \dfrac{dy}{dx} = -\sin x$$

Factor out $$\dfrac{dy}{dx}$$:

$$(1 + \cos y)\dfrac{dy}{dx} = -\sin x$$

$$\dfrac{dy}{dx} = \dfrac{-\sin x}{1 + \cos y}, \qquad 1 + \cos y \ne 0$$

Let $$y = \sin^{-1} x$$. Then $$x = \sin y$$, where $$y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$.

Differentiate $$x = \sin y$$ with respect to $$x$$:

$$1 = \cos y \cdot \dfrac{dy}{dx} \quad\Rightarrow\quad \dfrac{dy}{dx} = \dfrac{1}{\cos y}$$

For $$y \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ we have $$\cos y > 0$$, so $$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$.

Hence

$$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

Differentiate $$2x + 3y = \sin x$$ with respect to $$x$$:

$$2 + 3\dfrac{dy}{dx} = \cos x$$

$$3\dfrac{dy}{dx} = \cos x - 2$$

$$\dfrac{dy}{dx} = \dfrac{\cos x - 2}{3}$$

Differentiate $$2x + 3y = \sin y$$ with respect to $$x$$:

$$2 + 3\dfrac{dy}{dx} = \cos y \cdot \dfrac{dy}{dx}$$

Bring the $$\dfrac{dy}{dx}$$ terms together:

$$2 = \cos y \cdot \dfrac{dy}{dx} - 3\dfrac{dy}{dx} = (\cos y - 3)\dfrac{dy}{dx}$$

$$\dfrac{dy}{dx} = \dfrac{2}{\cos y - 3}$$

Differentiate $$ax + by^2 = \cos y$$ with respect to $$x$$:

$$a + 2by\dfrac{dy}{dx} = -\sin y \cdot \dfrac{dy}{dx}$$

Collect the $$\dfrac{dy}{dx}$$ terms:

$$a = -\sin y \cdot \dfrac{dy}{dx} - 2by\dfrac{dy}{dx} = -(2by + \sin y)\dfrac{dy}{dx}$$

$$\dfrac{dy}{dx} = \dfrac{-a}{2by + \sin y}$$

Differentiate $$xy + y^2 = \tan x + y$$ with respect to $$x$$, using the product rule on $$xy$$:

$$\left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = \sec^2 x + \dfrac{dy}{dx}$$

Collect the $$\dfrac{dy}{dx}$$ terms on one side:

$$x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} - \dfrac{dy}{dx} = \sec^2 x - y$$

$$(x + 2y - 1)\dfrac{dy}{dx} = \sec^2 x - y$$

$$\dfrac{dy}{dx} = \dfrac{\sec^2 x - y}{x + 2y - 1}$$

Differentiate $$x^2 + xy + y^2 = 100$$ with respect to $$x$$, using the product rule on $$xy$$:

$$2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$$

$$2x + y + (x + 2y)\dfrac{dy}{dx} = 0$$

$$\dfrac{dy}{dx} = \dfrac{-(2x + y)}{x + 2y}$$

Differentiate $$x^3 + x^2 y + xy^2 + y^3 = 81$$ with respect to $$x$$, using the product rule on $$x^2 y$$ and $$xy^2$$:

$$3x^2 + \left(2xy + x^2\dfrac{dy}{dx}\right) + \left(y^2 + 2xy\dfrac{dy}{dx}\right) + 3y^2\dfrac{dy}{dx} = 0$$

Group the terms containing $$\dfrac{dy}{dx}$$:

$$3x^2 + 2xy + y^2 + (x^2 + 2xy + 3y^2)\dfrac{dy}{dx} = 0$$

$$\dfrac{dy}{dx} = \dfrac{-(3x^2 + 2xy + y^2)}{x^2 + 2xy + 3y^2}$$

Differentiate $$\sin^2 y + \cos(xy) = \kappa$$ with respect to $$x$$ ($$\kappa$$ is a constant).

$$\dfrac{d}{dx}\sin^2 y = 2\sin y \cos y \cdot \dfrac{dy}{dx} = \sin 2y \cdot \dfrac{dy}{dx}$$

$$\dfrac{d}{dx}\cos(xy) = -\sin(xy) \cdot \dfrac{d}{dx}(xy) = -\sin(xy)\left(y + x\dfrac{dy}{dx}\right)$$

So the differentiated equation is

$$\sin 2y \cdot \dfrac{dy}{dx} - \sin(xy)\left(y + x\dfrac{dy}{dx}\right) = 0$$

$$\left(\sin 2y - x\sin(xy)\right)\dfrac{dy}{dx} = y\sin(xy)$$

$$\dfrac{dy}{dx} = \dfrac{y\sin(xy)}{\sin 2y - x\sin(xy)}$$

Differentiate $$\sin^2 x + \cos^2 y = 1$$ with respect to $$x$$:

$$2\sin x \cos x + 2\cos y(-\sin y)\dfrac{dy}{dx} = 0$$

Using $$2\sin x\cos x = \sin 2x$$ and $$2\sin y\cos y = \sin 2y$$,

$$\sin 2x - \sin 2y \cdot \dfrac{dy}{dx} = 0$$

$$\dfrac{dy}{dx} = \dfrac{\sin 2x}{\sin 2y}$$

Let $$y = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$$. Substitute $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$.

Then $$\dfrac{2x}{1 + x^2} = \dfrac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta$$.

So $$y = \sin^{-1}(\sin 2\theta)$$. For $$-1 < x < 1$$ we have $$\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$$, hence $$2\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ and $$y = 2\theta = 2\tan^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + x^2} = \dfrac{2}{1 + x^2}$$

Let $$y = \tan^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right)$$. Substitute $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$.

Then $$\dfrac{3x - x^3}{1 - 3x^2} = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta$$.

So $$y = \tan^{-1}(\tan 3\theta)$$. For $$-\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}$$ we have $$\theta \in \left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right)$$, hence $$3\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ and $$y = 3\theta = 3\tan^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 3 \cdot \dfrac{1}{1 + x^2} = \dfrac{3}{1 + x^2}$$

Let $$y = \cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right)$$. Substitute $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$.

Then $$\dfrac{1 - x^2}{1 + x^2} = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$$.

So $$y = \cos^{-1}(\cos 2\theta)$$. For $$0 < x < 1$$ we have $$\theta \in \left(0, \dfrac{\pi}{4}\right)$$, hence $$2\theta \in \left(0, \dfrac{\pi}{2}\right)$$ and $$y = 2\theta = 2\tan^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + x^2} = \dfrac{2}{1 + x^2}$$

Let $$y = \sin^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right)$$. Substitute $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$.

Then $$\dfrac{1 - x^2}{1 + x^2} = \cos 2\theta = \sin\left(\dfrac{\pi}{2} - 2\theta\right)$$.

So $$y = \sin^{-1}\left(\sin\left(\dfrac{\pi}{2} - 2\theta\right)\right)$$. For $$0 < x < 1$$, $$\theta \in \left(0, \dfrac{\pi}{4}\right)$$, so $$\dfrac{\pi}{2} - 2\theta \in \left(0, \dfrac{\pi}{2}\right) \subset \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$. Hence $$y = \dfrac{\pi}{2} - 2\theta = \dfrac{\pi}{2} - 2\tan^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 0 - 2 \cdot \dfrac{1}{1 + x^2} = \dfrac{-2}{1 + x^2}$$

Let $$y = \cos^{-1}\left(\dfrac{2x}{1 + x^2}\right)$$. Substitute $$x = \tan\theta$$, so that $$\theta = \tan^{-1} x$$.

Then $$\dfrac{2x}{1 + x^2} = \sin 2\theta = \cos\left(\dfrac{\pi}{2} - 2\theta\right)$$.

So $$y = \cos^{-1}\left(\cos\left(\dfrac{\pi}{2} - 2\theta\right)\right)$$. For $$-1 < x < 1$$, $$\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$$, so $$\dfrac{\pi}{2} - 2\theta \in (0, \pi)$$, the principal range of $$\cos^{-1}$$. Hence $$y = \dfrac{\pi}{2} - 2\theta = \dfrac{\pi}{2} - 2\tan^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = -2 \cdot \dfrac{1}{1 + x^2} = \dfrac{-2}{1 + x^2}$$

Let $$y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$$. Substitute $$x = \sin\theta$$, so that $$\theta = \sin^{-1} x$$.

Then $$2x\sqrt{1 - x^2} = 2\sin\theta\sqrt{1 - \sin^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta$$ (here $$\cos\theta \ge 0$$).

So $$y = \sin^{-1}(\sin 2\theta)$$. For $$-\dfrac{1}{\sqrt{2}} < x < \dfrac{1}{\sqrt{2}}$$, $$\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$$, so $$2\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$ and $$y = 2\theta = 2\sin^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{\sqrt{1 - x^2}} = \dfrac{2}{\sqrt{1 - x^2}}$$

Let $$y = \sec^{-1}\left(\dfrac{1}{2x^2 - 1}\right)$$. Substitute $$x = \cos\theta$$, so that $$\theta = \cos^{-1} x$$.

Then $$2x^2 - 1 = 2\cos^2\theta - 1 = \cos 2\theta$$, so $$\dfrac{1}{2x^2 - 1} = \dfrac{1}{\cos 2\theta} = \sec 2\theta$$.

So $$y = \sec^{-1}(\sec 2\theta)$$. For $$0 < x < \dfrac{1}{\sqrt{2}}$$, $$\theta \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$$, so $$2\theta \in \left(\dfrac{\pi}{2}, \pi\right)$$, which lies in the principal range of $$\sec^{-1}$$. Hence $$y = 2\theta = 2\cos^{-1} x$$.

Differentiating,

$$\dfrac{dy}{dx} = 2 \cdot \left(\dfrac{-1}{\sqrt{1 - x^2}}\right) = \dfrac{-2}{\sqrt{1 - x^2}}$$

The natural logarithm $$\log x$$ is defined only for $$x > 0$$. Therefore the expression $$e^{\log x}$$ itself only makes sense when $$x > 0$$.

For $$x > 0$$: if $$y = \log x$$, then by the definition of the logarithm $$x = e^y = e^{\log x}$$. So the identity $$x = e^{\log x}$$ holds for every $$x > 0$$.

For $$x \le 0$$: $$\log x$$ is not defined, hence $$e^{\log x}$$ is not defined, and the statement cannot hold.

Hence it is not true that $$x = e^{\log x}$$ for all real $$x$$; the identity is valid only for $$x > 0$$.

Let $$y = \dfrac{e^x}{\sin x}$$. Apply the quotient rule with $$u = e^x$$ and $$v = \sin x$$, so $$u' = e^x$$ and $$v' = \cos x$$.

$$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{e^x \sin x - e^x \cos x}{\sin^2 x}$$

$$\dfrac{dy}{dx} = \dfrac{e^x(\sin x - \cos x)}{\sin^2 x}, \qquad x \ne n\pi,\ n \in \mathbf{Z}$$

Let $$y = e^{\sin^{-1} x}$$. By the chain rule, with inner function $$u = \sin^{-1} x$$,

$$\dfrac{dy}{dx} = e^{\sin^{-1} x} \cdot \dfrac{d}{dx}(\sin^{-1} x) = e^{\sin^{-1} x} \cdot \dfrac{1}{\sqrt{1 - x^2}}$$

$$\dfrac{dy}{dx} = \dfrac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

Let $$y = e^{x^3}$$. By the chain rule, with inner function $$u = x^3$$,

$$\dfrac{dy}{dx} = e^{x^3} \cdot \dfrac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2$$

$$\dfrac{dy}{dx} = 3x^2 e^{x^3}$$

Let $$y = \sin(\tan^{-1} e^{-x})$$. Differentiate as a composition, from the outside inward.

$$\dfrac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \dfrac{d}{dx}(\tan^{-1} e^{-x})$$

$$\dfrac{d}{dx}(\tan^{-1} e^{-x}) = \dfrac{1}{1 + (e^{-x})^2} \cdot \dfrac{d}{dx}(e^{-x}) = \dfrac{-e^{-x}}{1 + e^{-2x}}$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{-e^{-x}\cos(\tan^{-1} e^{-x})}{1 + e^{-2x}}$$

Let $$y = \log(\cos e^x)$$. Differentiate as a composition.

$$\dfrac{dy}{dx} = \dfrac{1}{\cos e^x} \cdot \dfrac{d}{dx}(\cos e^x)$$

$$\dfrac{d}{dx}(\cos e^x) = -\sin e^x \cdot \dfrac{d}{dx}(e^x) = -e^x \sin e^x$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{-e^x \sin e^x}{\cos e^x} = -e^x \tan e^x$$

Let $$y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$$. Differentiate term by term, using the chain rule on each term.

For each $$n$$, $$\dfrac{d}{dx}e^{x^n} = e^{x^n} \cdot \dfrac{d}{dx}(x^n) = n x^{n-1} e^{x^n}$$.

Hence

$$\dfrac{dy}{dx} = e^x + 2x\,e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}$$

Let $$y = \sqrt{e^{\sqrt{x}}}$$, $$x > 0$$. Differentiate as a composition.

$$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot \dfrac{d}{dx}\left(e^{\sqrt{x}}\right)$$

$$\dfrac{d}{dx}\left(e^{\sqrt{x}}\right) = e^{\sqrt{x}} \cdot \dfrac{d}{dx}\sqrt{x} = e^{\sqrt{x}} \cdot \dfrac{1}{2\sqrt{x}}$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot \dfrac{e^{\sqrt{x}}}{2\sqrt{x}} = \dfrac{e^{\sqrt{x}}}{4\sqrt{x}\,\sqrt{e^{\sqrt{x}}}} = \dfrac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$$

Let $$y = \log(\log x)$$, $$x > 1$$. By the chain rule, with inner function $$u = \log x$$,

$$\dfrac{dy}{dx} = \dfrac{1}{\log x} \cdot \dfrac{d}{dx}(\log x) = \dfrac{1}{\log x} \cdot \dfrac{1}{x}$$

$$\dfrac{dy}{dx} = \dfrac{1}{x\log x}$$

Let $$y = \dfrac{\cos x}{\log x}$$. Apply the quotient rule with $$u = \cos x$$ and $$v = \log x$$, so $$u' = -\sin x$$ and $$v' = \dfrac{1}{x}$$.

$$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{(-\sin x)(\log x) - (\cos x)\left(\dfrac{1}{x}\right)}{(\log x)^2}$$

Multiplying numerator and denominator by $$x$$,

$$\dfrac{dy}{dx} = \dfrac{-\left(x\sin x\,\log x + \cos x\right)}{x(\log x)^2}$$

Let $$y = \cos(\log x + e^x)$$, $$x > 0$$. By the chain rule, with inner function $$u = \log x + e^x$$,

$$\dfrac{dy}{dx} = -\sin(\log x + e^x) \cdot \dfrac{d}{dx}(\log x + e^x)$$

$$\dfrac{d}{dx}(\log x + e^x) = \dfrac{1}{x} + e^x$$

Therefore

$$\dfrac{dy}{dx} = -\left(\dfrac{1}{x} + e^x\right)\sin(\log x + e^x)$$

Let $$y = \sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}$$. Since this is a product and quotient under a square root, use logarithmic differentiation.

Take the logarithm of both sides:

$$\log y = \dfrac{1}{2}\Big[\log(x - 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5)\Big]$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{2}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]$$

Hence

$$\dfrac{dy}{dx} = \dfrac{1}{2}\sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]$$

Let $$y = a^x$$, where $$a > 0$$. Take the logarithm of both sides:

$$\log y = x \log a$$

Differentiate both sides with respect to $$x$$ (note $$\log a$$ is a constant):

$$\dfrac{1}{y}\dfrac{dy}{dx} = \log a$$

$$\dfrac{dy}{dx} = y \log a = a^x \log a$$

Equivalently, writing $$a^x = e^{x\log a}$$ and using the chain rule gives the same result, $$\dfrac{dy}{dx} = e^{x\log a}\cdot\log a = a^x \log a$$.

Let $$y = x^{\sin x}$$, $$x > 0$$. Since both the base and the exponent contain $$x$$, use logarithmic differentiation.

Take the logarithm of both sides: $$\log y = \sin x \cdot \log x$$.

Differentiate both sides with respect to $$x$$, using the product rule on the right:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \cos x \cdot \log x + \sin x \cdot \dfrac{1}{x}$$

$$\dfrac{dy}{dx} = y\left(\cos x \log x + \dfrac{\sin x}{x}\right) = x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)$$

Let $$u = y^x$$, $$v = x^y$$ and $$w = x^x$$, so that $$u + v + w = a^b$$. Since $$a^b$$ is a constant,

$$\dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx} = 0 \qquad \text{...(1)}$$

For $$u = y^x$$: $$\log u = x\log y$$, so $$\dfrac{1}{u}\dfrac{du}{dx} = \log y + \dfrac{x}{y}\dfrac{dy}{dx}$$, giving $$\dfrac{du}{dx} = y^x\left(\log y + \dfrac{x}{y}\dfrac{dy}{dx}\right)$$.

For $$v = x^y$$: $$\log v = y\log x$$, so $$\dfrac{1}{v}\dfrac{dv}{dx} = \dfrac{dy}{dx}\log x + \dfrac{y}{x}$$, giving $$\dfrac{dv}{dx} = x^y\left(\dfrac{dy}{dx}\log x + \dfrac{y}{x}\right)$$.

For $$w = x^x$$: $$\log w = x\log x$$, so $$\dfrac{1}{w}\dfrac{dw}{dx} = \log x + 1$$, giving $$\dfrac{dw}{dx} = x^x(1 + \log x)$$.

Substitute these into (1):

$$y^x\left(\log y + \dfrac{x}{y}\dfrac{dy}{dx}\right) + x^y\left(\dfrac{dy}{dx}\log x + \dfrac{y}{x}\right) + x^x(1 + \log x) = 0$$

Collecting the terms containing $$\dfrac{dy}{dx}$$ (note $$\dfrac{x}{y}y^x = x\,y^{x-1}$$ and $$\dfrac{y}{x}x^y = y\,x^{y-1}$$):

$$\left(x\,y^{x-1} + x^y\log x\right)\dfrac{dy}{dx} = -\left(y^x\log y + y\,x^{y-1} + x^x(1 + \log x)\right)$$

$$\dfrac{dy}{dx} = \dfrac{-\left[y^x\log y + y\,x^{y-1} + x^x(1 + \log x)\right]}{x\,y^{x-1} + x^y\log x}$$

Let $$y = \cos x \cdot \cos 2x \cdot \cos 3x$$. Use logarithmic differentiation.

$$\log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x)$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{-\sin x}{\cos x} + \dfrac{-2\sin 2x}{\cos 2x} + \dfrac{-3\sin 3x}{\cos 3x}$$

$$\dfrac{1}{y}\dfrac{dy}{dx} = -(\tan x + 2\tan 2x + 3\tan 3x)$$

$$\dfrac{dy}{dx} = -\cos x\,\cos 2x\,\cos 3x\,(\tan x + 2\tan 2x + 3\tan 3x)$$

Let $$y = \sqrt{\dfrac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)}}$$. Use logarithmic differentiation.

$$\log y = \dfrac{1}{2}\Big[\log(x - 1) + \log(x - 2) - \log(x - 3) - \log(x - 4) - \log(x - 5)\Big]$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{2}\left[\dfrac{1}{x - 1} + \dfrac{1}{x - 2} - \dfrac{1}{x - 3} - \dfrac{1}{x - 4} - \dfrac{1}{x - 5}\right]$$

Hence

$$\dfrac{dy}{dx} = \dfrac{1}{2}\sqrt{\dfrac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)}}\left[\dfrac{1}{x - 1} + \dfrac{1}{x - 2} - \dfrac{1}{x - 3} - \dfrac{1}{x - 4} - \dfrac{1}{x - 5}\right]$$

Let $$y = (\log x)^{\cos x}$$. Since the base and the exponent both involve $$x$$, use logarithmic differentiation.

$$\log y = \cos x \cdot \log(\log x)$$

Differentiate both sides with respect to $$x$$, using the product rule on the right:

$$\dfrac{1}{y}\dfrac{dy}{dx} = -\sin x \cdot \log(\log x) + \cos x \cdot \dfrac{1}{\log x} \cdot \dfrac{1}{x}$$

$$\dfrac{dy}{dx} = (\log x)^{\cos x}\left[\dfrac{\cos x}{x\log x} - \sin x\,\log(\log x)\right]$$

Let $$y = x^x - 2^{\sin x}$$. Write $$y = u - v$$, where $$u = x^x$$ and $$v = 2^{\sin x}$$, and differentiate each by logarithms.

$$u = x^x$$: $$\log u = x\log x$$, so $$\dfrac{1}{u}\dfrac{du}{dx} = \log x + 1$$, giving $$\dfrac{du}{dx} = x^x(1 + \log x)$$.

$$v = 2^{\sin x}$$: $$\log v = \sin x \cdot \log 2$$, so $$\dfrac{1}{v}\dfrac{dv}{dx} = \cos x \cdot \log 2$$, giving $$\dfrac{dv}{dx} = 2^{\sin x}\cos x\,\log 2$$.

Therefore

$$\dfrac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x}\cos x\,\log 2$$

Let $$y = (x + 3)^2 (x + 4)^3 (x + 5)^4$$. Use logarithmic differentiation.

$$\log y = 2\log(x + 3) + 3\log(x + 4) + 4\log(x + 5)$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{2}{x + 3} + \dfrac{3}{x + 4} + \dfrac{4}{x + 5}$$

$$\dfrac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4\left[\dfrac{2}{x + 3} + \dfrac{3}{x + 4} + \dfrac{4}{x + 5}\right]$$

Taking out the common factors $$(x + 3)(x + 4)^2(x + 5)^3$$, the bracket becomes $$2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4) = 9x^2 + 70x + 133$$, so

$$\dfrac{dy}{dx} = (x + 3)(x + 4)^2 (x + 5)^3\,(9x^2 + 70x + 133)$$

Let $$y = u + v$$, where $$u = \left(x + \dfrac{1}{x}\right)^x$$ and $$v = x^{1 + \frac{1}{x}}$$. Differentiate each by logarithms.

$$u = \left(x + \dfrac{1}{x}\right)^x$$: $$\log u = x\log\left(x + \dfrac{1}{x}\right)$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = \log\left(x + \dfrac{1}{x}\right) + x \cdot \dfrac{1 - \frac{1}{x^2}}{x + \frac{1}{x}}$$

Since $$x + \dfrac{1}{x} = \dfrac{x^2 + 1}{x}$$ and $$1 - \dfrac{1}{x^2} = \dfrac{x^2 - 1}{x^2}$$, the last term simplifies to $$\dfrac{x^2 - 1}{x^2 + 1}$$. Hence

$$\dfrac{du}{dx} = \left(x + \dfrac{1}{x}\right)^x\left[\log\left(x + \dfrac{1}{x}\right) + \dfrac{x^2 - 1}{x^2 + 1}\right]$$

$$v = x^{1 + \frac{1}{x}}$$: $$\log v = \left(1 + \dfrac{1}{x}\right)\log x$$.

$$\dfrac{1}{v}\dfrac{dv}{dx} = -\dfrac{1}{x^2}\log x + \left(1 + \dfrac{1}{x}\right)\dfrac{1}{x} = \dfrac{x + 1 - \log x}{x^2}$$

$$\dfrac{dv}{dx} = x^{1 + \frac{1}{x}}\cdot\dfrac{x + 1 - \log x}{x^2}$$

Therefore

$$\dfrac{dy}{dx} = \left(x + \dfrac{1}{x}\right)^x\left[\log\left(x + \dfrac{1}{x}\right) + \dfrac{x^2 - 1}{x^2 + 1}\right] + x^{1 + \frac{1}{x}}\cdot\dfrac{x + 1 - \log x}{x^2}$$

Let $$y = u + v$$, where $$u = (\log x)^x$$ and $$v = x^{\log x}$$.

$$u = (\log x)^x$$: $$\log u = x\log(\log x)$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = \log(\log x) + x \cdot \dfrac{1}{\log x}\cdot\dfrac{1}{x} = \log(\log x) + \dfrac{1}{\log x}$$

$$\dfrac{du}{dx} = (\log x)^x\left[\log(\log x) + \dfrac{1}{\log x}\right]$$

$$v = x^{\log x}$$: $$\log v = \log x \cdot \log x = (\log x)^2$$.

$$\dfrac{1}{v}\dfrac{dv}{dx} = 2\log x \cdot \dfrac{1}{x}$$

$$\dfrac{dv}{dx} = x^{\log x}\cdot\dfrac{2\log x}{x}$$

Therefore

$$\dfrac{dy}{dx} = (\log x)^x\left[\log(\log x) + \dfrac{1}{\log x}\right] + \dfrac{2\,x^{\log x}\log x}{x}$$

Let $$y = u + v$$, where $$u = (\sin x)^x$$ and $$v = \sin^{-1}\sqrt{x}$$.

$$u = (\sin x)^x$$: $$\log u = x\log(\sin x)$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = \log(\sin x) + x \cdot \dfrac{\cos x}{\sin x} = \log(\sin x) + x\cot x$$

$$\dfrac{du}{dx} = (\sin x)^x\left[\log(\sin x) + x\cot x\right]$$

$$v = \sin^{-1}\sqrt{x}$$: by the chain rule,

$$\dfrac{dv}{dx} = \dfrac{1}{\sqrt{1 - (\sqrt{x})^2}}\cdot\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{\sqrt{1 - x}}\cdot\dfrac{1}{2\sqrt{x}} = \dfrac{1}{2\sqrt{x - x^2}}$$

Therefore

$$\dfrac{dy}{dx} = (\sin x)^x\left[\log(\sin x) + x\cot x\right] + \dfrac{1}{2\sqrt{x - x^2}}$$

Let $$y = u + v$$, where $$u = x^{\sin x}$$ and $$v = (\sin x)^{\cos x}$$.

$$u = x^{\sin x}$$: $$\log u = \sin x\,\log x$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = \cos x\,\log x + \dfrac{\sin x}{x}$$

$$\dfrac{du}{dx} = x^{\sin x}\left(\cos x\,\log x + \dfrac{\sin x}{x}\right)$$

$$v = (\sin x)^{\cos x}$$: $$\log v = \cos x\,\log(\sin x)$$.

$$\dfrac{1}{v}\dfrac{dv}{dx} = -\sin x\,\log(\sin x) + \cos x \cdot \dfrac{\cos x}{\sin x} = -\sin x\,\log(\sin x) + \cos x\cot x$$

$$\dfrac{dv}{dx} = (\sin x)^{\cos x}\left[\cos x\cot x - \sin x\,\log(\sin x)\right]$$

Therefore

$$\dfrac{dy}{dx} = x^{\sin x}\left(\cos x\,\log x + \dfrac{\sin x}{x}\right) + (\sin x)^{\cos x}\left[\cos x\cot x - \sin x\,\log(\sin x)\right]$$

Let $$y = u + v$$, where $$u = x^{x\cos x}$$ and $$v = \dfrac{x^2 + 1}{x^2 - 1}$$.

$$u = x^{x\cos x}$$: $$\log u = x\cos x\,\log x$$. Differentiate the right side as a product of three functions:

$$\dfrac{1}{u}\dfrac{du}{dx} = \cos x\,\log x + x(-\sin x)\log x + x\cos x\cdot\dfrac{1}{x}$$

$$\dfrac{1}{u}\dfrac{du}{dx} = \cos x\,\log x - x\sin x\,\log x + \cos x$$

$$\dfrac{du}{dx} = x^{x\cos x}\left[\cos x(1 + \log x) - x\sin x\,\log x\right]$$

$$v = \dfrac{x^2 + 1}{x^2 - 1}$$: by the quotient rule,

$$\dfrac{dv}{dx} = \dfrac{2x(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \dfrac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} = \dfrac{-4x}{(x^2 - 1)^2}$$

Therefore

$$\dfrac{dy}{dx} = x^{x\cos x}\left[\cos x(1 + \log x) - x\sin x\,\log x\right] - \dfrac{4x}{(x^2 - 1)^2}$$

Let $$y = u + v$$, where $$u = (x\cos x)^x$$ and $$v = (x\sin x)^{1/x}$$.

$$u = (x\cos x)^x$$: $$\log u = x\log(x\cos x) = x[\log x + \log\cos x]$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = \log(x\cos x) + x\left[\dfrac{1}{x} - \tan x\right] = \log(x\cos x) + 1 - x\tan x$$

$$\dfrac{du}{dx} = (x\cos x)^x\left[\log(x\cos x) + 1 - x\tan x\right]$$

$$v = (x\sin x)^{1/x}$$: $$\log v = \dfrac{1}{x}\log(x\sin x)$$. Using $$\dfrac{d}{dx}\log(x\sin x) = \dfrac{1}{x} + \cot x$$,

$$\dfrac{1}{v}\dfrac{dv}{dx} = -\dfrac{1}{x^2}\log(x\sin x) + \dfrac{1}{x}\left(\dfrac{1}{x} + \cot x\right) = \dfrac{1 + x\cot x - \log(x\sin x)}{x^2}$$

$$\dfrac{dv}{dx} = (x\sin x)^{1/x}\cdot\dfrac{1 + x\cot x - \log(x\sin x)}{x^2}$$

Therefore

$$\dfrac{dy}{dx} = (x\cos x)^x\left[\log(x\cos x) + 1 - x\tan x\right] + (x\sin x)^{1/x}\cdot\dfrac{1 + x\cot x - \log(x\sin x)}{x^2}$$

Let $$u = x^y$$ and $$v = y^x$$, so $$u + v = 1$$ and hence $$\dfrac{du}{dx} + \dfrac{dv}{dx} = 0$$.

$$u = x^y$$: $$\log u = y\log x$$, so $$\dfrac{1}{u}\dfrac{du}{dx} = \dfrac{dy}{dx}\log x + \dfrac{y}{x}$$, giving $$\dfrac{du}{dx} = x^y\left(\dfrac{dy}{dx}\log x + \dfrac{y}{x}\right)$$.

$$v = y^x$$: $$\log v = x\log y$$, so $$\dfrac{1}{v}\dfrac{dv}{dx} = \log y + \dfrac{x}{y}\dfrac{dy}{dx}$$, giving $$\dfrac{dv}{dx} = y^x\left(\log y + \dfrac{x}{y}\dfrac{dy}{dx}\right)$$.

Substituting into $$\dfrac{du}{dx} + \dfrac{dv}{dx} = 0$$:

$$x^y\left(\dfrac{dy}{dx}\log x + \dfrac{y}{x}\right) + y^x\left(\log y + \dfrac{x}{y}\dfrac{dy}{dx}\right) = 0$$

Collect the $$\dfrac{dy}{dx}$$ terms (note $$\dfrac{x}{y}y^x = x\,y^{x-1}$$ and $$\dfrac{y}{x}x^y = y\,x^{y-1}$$):

$$\left(x^y\log x + x\,y^{x-1}\right)\dfrac{dy}{dx} = -\left(y\,x^{y-1} + y^x\log y\right)$$

$$\dfrac{dy}{dx} = \dfrac{-\left(y\,x^{y-1} + y^x\log y\right)}{x^y\log x + x\,y^{x-1}}$$

Given $$y^x = x^y$$. Take the logarithm of both sides:

$$x\log y = y\log x$$

Differentiate both sides with respect to $$x$$, using the product rule on each side:

$$\log y + x \cdot \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{dy}{dx}\log x + y \cdot \dfrac{1}{x}$$

Collect the $$\dfrac{dy}{dx}$$ terms on one side:

$$\dfrac{x}{y}\dfrac{dy}{dx} - \log x\,\dfrac{dy}{dx} = \dfrac{y}{x} - \log y$$

$$\left(\dfrac{x}{y} - \log x\right)\dfrac{dy}{dx} = \dfrac{y}{x} - \log y$$

Multiplying numerator and denominator by $$xy$$,

$$\dfrac{dy}{dx} = \dfrac{y^2 - xy\log y}{x^2 - xy\log x} = \dfrac{y(y - x\log y)}{x(x - y\log x)}$$

Given $$(\cos x)^y = (\cos y)^x$$. Take the logarithm of both sides:

$$y\log(\cos x) = x\log(\cos y)$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{dy}{dx}\log(\cos x) + y\cdot\dfrac{-\sin x}{\cos x} = \log(\cos y) + x\cdot\dfrac{-\sin y}{\cos y}\dfrac{dy}{dx}$$

$$\dfrac{dy}{dx}\log(\cos x) - y\tan x = \log(\cos y) - x\tan y\,\dfrac{dy}{dx}$$

Collect the $$\dfrac{dy}{dx}$$ terms:

$$\left(\log(\cos x) + x\tan y\right)\dfrac{dy}{dx} = \log(\cos y) + y\tan x$$

$$\dfrac{dy}{dx} = \dfrac{\log(\cos y) + y\tan x}{\log(\cos x) + x\tan y}$$

Given $$xy = e^{x - y}$$. Take the logarithm of both sides:

$$\log x + \log y = x - y$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} = 1 - \dfrac{dy}{dx}$$

Collect the $$\dfrac{dy}{dx}$$ terms:

$$\dfrac{1}{y}\dfrac{dy}{dx} + \dfrac{dy}{dx} = 1 - \dfrac{1}{x}$$

$$\left(\dfrac{1 + y}{y}\right)\dfrac{dy}{dx} = \dfrac{x - 1}{x}$$

$$\dfrac{dy}{dx} = \dfrac{y(x - 1)}{x(1 + y)}$$

Let $$f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)$$. Use logarithmic differentiation.

$$\log f(x) = \log(1 + x) + \log(1 + x^2) + \log(1 + x^4) + \log(1 + x^8)$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{f'(x)}{f(x)} = \dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}$$

$$f'(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]$$

Finding $$f'(1)$$: At $$x = 1$$, $$f(1) = (2)(2)(2)(2) = 16$$, and the bracket becomes

$$\dfrac{1}{2} + \dfrac{2}{2} + \dfrac{4}{2} + \dfrac{8}{2} = \dfrac{1}{2} + 1 + 2 + 4 = \dfrac{15}{2}$$

Therefore $$f'(1) = 16 \times \dfrac{15}{2} = 120$$.

Let $$y = u \cdot v \cdot w$$, where $$u$$, $$v$$, $$w$$ are functions of $$x$$.

Method 1 — repeated product rule. Group the product as $$y = (u \cdot v) \cdot w$$. By the product rule,

$$\dfrac{dy}{dx} = \dfrac{d(uv)}{dx}\cdot w + (uv)\cdot\dfrac{dw}{dx}$$

Again by the product rule, $$\dfrac{d(uv)}{dx} = \dfrac{du}{dx}v + u\dfrac{dv}{dx}$$. Substituting,

$$\dfrac{dy}{dx} = \left(\dfrac{du}{dx}v + u\dfrac{dv}{dx}\right)w + uv\dfrac{dw}{dx} = \dfrac{du}{dx}v\,w + u\dfrac{dv}{dx}w + uv\dfrac{dw}{dx}$$

Method 2 — logarithmic differentiation. Assuming $$u, v, w > 0$$, take logarithms:

$$\log y = \log u + \log v + \log w$$

Differentiate both sides with respect to $$x$$:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{u}\dfrac{du}{dx} + \dfrac{1}{v}\dfrac{dv}{dx} + \dfrac{1}{w}\dfrac{dw}{dx}$$

Multiply both sides by $$y = uvw$$:

$$\dfrac{dy}{dx} = uvw\left(\dfrac{1}{u}\dfrac{du}{dx} + \dfrac{1}{v}\dfrac{dv}{dx} + \dfrac{1}{w}\dfrac{dw}{dx}\right) = \dfrac{du}{dx}v\,w + u\dfrac{dv}{dx}w + uv\dfrac{dw}{dx}$$

Both methods yield the same result, which proves the formula.

Here $$x$$ and $$y$$ are given in terms of the parameter $$\theta$$. Differentiate each with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = -a\sin\theta, \qquad \dfrac{dy}{d\theta} = a\cos\theta$$

Then

$$\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{a\cos\theta}{-a\sin\theta} = -\cot\theta$$

Differentiate $$x$$ and $$y$$ with respect to the parameter $$t$$:

$$\dfrac{dx}{dt} = 2at, \qquad \dfrac{dy}{dt} = 2a$$

Then

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2a}{2at} = \dfrac{1}{t}$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = a(1 + \cos\theta), \qquad \dfrac{dy}{d\theta} = a\sin\theta$$

Then

$$\dfrac{dy}{dx} = \dfrac{a\sin\theta}{a(1 + \cos\theta)} = \dfrac{\sin\theta}{1 + \cos\theta}$$

Using $$\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$$ and $$1 + \cos\theta = 2\cos^2\dfrac{\theta}{2}$$,

$$\dfrac{dy}{dx} = \dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \tan\dfrac{\theta}{2}$$

Differentiate $$x^{2/3} + y^{2/3} = a^{2/3}$$ with respect to $$x$$ (here $$a$$ is a constant):

$$\dfrac{2}{3}x^{-1/3} + \dfrac{2}{3}y^{-1/3}\dfrac{dy}{dx} = 0$$

$$\dfrac{2}{3}y^{-1/3}\dfrac{dy}{dx} = -\dfrac{2}{3}x^{-1/3}$$

$$\dfrac{dy}{dx} = -\dfrac{x^{-1/3}}{y^{-1/3}} = -\dfrac{y^{1/3}}{x^{1/3}} = -\left(\dfrac{y}{x}\right)^{1/3}$$

Differentiate $$x$$ and $$y$$ with respect to the parameter $$t$$:

$$\dfrac{dx}{dt} = 4at, \qquad \dfrac{dy}{dt} = 4at^3$$

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4at^3}{4at} = t^2$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = -a\sin\theta, \qquad \dfrac{dy}{d\theta} = -b\sin\theta$$

$$\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{-b\sin\theta}{-a\sin\theta} = \dfrac{b}{a}$$

Differentiate $$x$$ and $$y$$ with respect to $$t$$:

$$\dfrac{dx}{dt} = \cos t, \qquad \dfrac{dy}{dt} = -2\sin 2t$$

$$\dfrac{dy}{dx} = \dfrac{-2\sin 2t}{\cos t}$$

Using $$\sin 2t = 2\sin t\cos t$$,

$$\dfrac{dy}{dx} = \dfrac{-2(2\sin t\cos t)}{\cos t} = -4\sin t$$

Differentiate $$x$$ and $$y$$ with respect to $$t$$:

$$\dfrac{dx}{dt} = 4, \qquad \dfrac{dy}{dt} = \dfrac{d}{dt}(4t^{-1}) = -\dfrac{4}{t^2}$$

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-4/t^2}{4} = -\dfrac{1}{t^2}$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = -\sin\theta + 2\sin 2\theta$$

$$\dfrac{dy}{d\theta} = \cos\theta - 2\cos 2\theta$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta}$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = a(1 - \cos\theta), \qquad \dfrac{dy}{d\theta} = -a\sin\theta$$

$$\dfrac{dy}{dx} = \dfrac{-a\sin\theta}{a(1 - \cos\theta)} = \dfrac{-\sin\theta}{1 - \cos\theta}$$

Using $$\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$$ and $$1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$$,

$$\dfrac{dy}{dx} = \dfrac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = -\cot\dfrac{\theta}{2}$$

Write $$x = \sin^3 t\,(\cos 2t)^{-1/2}$$ and $$y = \cos^3 t\,(\cos 2t)^{-1/2}$$. Differentiate using the product and chain rules, with $$\dfrac{d}{dt}\cos 2t = -2\sin 2t$$.

Derivative of $$x$$:

$$\dfrac{dx}{dt} = 3\sin^2 t\cos t\,(\cos 2t)^{-1/2} + \sin^3 t\,\sin 2t\,(\cos 2t)^{-3/2}$$

$$= (\cos 2t)^{-3/2}\big[3\sin^2 t\cos t\cos 2t + \sin^3 t\sin 2t\big]$$

With $$\sin 2t = 2\sin t\cos t$$ and $$\cos 2t = 1 - 2\sin^2 t$$, the bracket equals $$\sin^2 t\cos t\,(3\cos 2t + 2\sin^2 t) = \sin^2 t\cos t\,(3 - 4\sin^2 t)$$. Hence

$$\dfrac{dx}{dt} = (\cos 2t)^{-3/2}\,\sin^2 t\cos t\,(3 - 4\sin^2 t)$$

Derivative of $$y$$: by the same working,

$$\dfrac{dy}{dt} = (\cos 2t)^{-3/2}\,\cos^2 t\sin t\,(3 - 4\cos^2 t)$$

Forming the ratio. Using $$\sin 3t = \sin t\,(3 - 4\sin^2 t)$$ and $$\cos 3t = -\cos t\,(3 - 4\cos^2 t)$$,

$$\dfrac{dy}{dx} = \dfrac{\cos^2 t\sin t\,(3 - 4\cos^2 t)}{\sin^2 t\cos t\,(3 - 4\sin^2 t)} = \dfrac{\cos^2 t\sin t\left(\dfrac{-\cos 3t}{\cos t}\right)}{\sin^2 t\cos t\left(\dfrac{\sin 3t}{\sin t}\right)}$$

$$= \dfrac{-\cos t\sin t\cos 3t}{\sin t\cos t\sin 3t} = -\dfrac{\cos 3t}{\sin 3t} = -\cot 3t$$

First compute $$\dfrac{d}{dt}\log\tan\dfrac{t}{2}$$:

$$\dfrac{d}{dt}\log\tan\dfrac{t}{2} = \dfrac{1}{\tan\frac{t}{2}}\cdot\sec^2\dfrac{t}{2}\cdot\dfrac{1}{2} = \dfrac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}} = \dfrac{1}{\sin t}$$

Hence, differentiating $$x$$ with respect to $$t$$,

$$\dfrac{dx}{dt} = a\left(-\sin t + \dfrac{1}{\sin t}\right) = a\cdot\dfrac{1 - \sin^2 t}{\sin t} = a\cdot\dfrac{\cos^2 t}{\sin t}$$

Also $$\dfrac{dy}{dt} = a\cos t$$.

Therefore

$$\dfrac{dy}{dx} = \dfrac{a\cos t}{a\cos^2 t/\sin t} = \dfrac{\sin t}{\cos t} = \tan t$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$:

$$\dfrac{dx}{d\theta} = a\sec\theta\tan\theta, \qquad \dfrac{dy}{d\theta} = b\sec^2\theta$$

$$\dfrac{dy}{dx} = \dfrac{b\sec^2\theta}{a\sec\theta\tan\theta} = \dfrac{b\sec\theta}{a\tan\theta}$$

Since $$\dfrac{\sec\theta}{\tan\theta} = \dfrac{1/\cos\theta}{\sin\theta/\cos\theta} = \dfrac{1}{\sin\theta} = \operatorname{cosec}\theta$$,

$$\dfrac{dy}{dx} = \dfrac{b}{a}\operatorname{cosec}\theta$$

Differentiate $$x$$ and $$y$$ with respect to $$\theta$$, using the product rule on $$\theta\sin\theta$$ and $$\theta\cos\theta$$:

$$\dfrac{dx}{d\theta} = a\big(-\sin\theta + \sin\theta + \theta\cos\theta\big) = a\theta\cos\theta$$

$$\dfrac{dy}{d\theta} = a\big(\cos\theta - \cos\theta + \theta\sin\theta\big) = a\theta\sin\theta$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta$$

Write $$x = \sqrt{a^{\sin^{-1} t}} = a^{\frac{1}{2}\sin^{-1} t}$$ and $$y = \sqrt{a^{\cos^{-1} t}} = a^{\frac{1}{2}\cos^{-1} t}$$.

Differentiate $$x$$: taking logarithms, $$\log x = \dfrac{1}{2}\sin^{-1} t\cdot\log a$$, so

$$\dfrac{1}{x}\dfrac{dx}{dt} = \dfrac{\log a}{2}\cdot\dfrac{1}{\sqrt{1 - t^2}} \quad\Rightarrow\quad \dfrac{dx}{dt} = \dfrac{x\log a}{2\sqrt{1 - t^2}}$$

Differentiate $$y$$: $$\log y = \dfrac{1}{2}\cos^{-1} t\cdot\log a$$, so

$$\dfrac{1}{y}\dfrac{dy}{dt} = \dfrac{\log a}{2}\cdot\dfrac{-1}{\sqrt{1 - t^2}} \quad\Rightarrow\quad \dfrac{dy}{dt} = \dfrac{-y\log a}{2\sqrt{1 - t^2}}$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-y\log a/(2\sqrt{1 - t^2})}{x\log a/(2\sqrt{1 - t^2})} = -\dfrac{y}{x}$$

Hence $$\dfrac{dy}{dx} = -\dfrac{y}{x}$$, as required.

Given $$y = x^3 + \tan x$$. Differentiate once with respect to $$x$$:

$$\dfrac{dy}{dx} = 3x^2 + \sec^2 x$$

Differentiate again. Note that $$\dfrac{d}{dx}\sec^2 x = 2\sec x\cdot(\sec x\tan x) = 2\sec^2 x\tan x$$.

$$\dfrac{d^2 y}{dx^2} = 6x + 2\sec^2 x\tan x$$

Given $$y = A\sin x + B\cos x$$. Differentiate with respect to $$x$$:

$$\dfrac{dy}{dx} = A\cos x - B\sin x$$

Differentiate again:

$$\dfrac{d^2 y}{dx^2} = -A\sin x - B\cos x = -(A\sin x + B\cos x) = -y$$

Therefore $$\dfrac{d^2 y}{dx^2} + y = 0$$, as required.

Given $$y = 3e^{2x} + 2e^{3x}$$.

$$\dfrac{dy}{dx} = 6e^{2x} + 6e^{3x}$$

$$\dfrac{d^2 y}{dx^2} = 12e^{2x} + 18e^{3x}$$

Substitute into the left side of the given equation:

$$\dfrac{d^2 y}{dx^2} - 5\dfrac{dy}{dx} + 6y = (12e^{2x} + 18e^{3x}) - 5(6e^{2x} + 6e^{3x}) + 6(3e^{2x} + 2e^{3x})$$

$$= (12 - 30 + 18)e^{2x} + (18 - 30 + 12)e^{3x} = 0\cdot e^{2x} + 0\cdot e^{3x} = 0$$

Hence $$\dfrac{d^2 y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$$.

Given $$y = \sin^{-1} x$$. Then

$$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$$

So $$\sqrt{1 - x^2}\,\dfrac{dy}{dx} = 1$$. Squaring both sides,

$$(1 - x^2)\left(\dfrac{dy}{dx}\right)^2 = 1$$

Differentiate both sides with respect to $$x$$:

$$(1 - x^2)\cdot 2\dfrac{dy}{dx}\dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2(-2x) = 0$$

Divide throughout by $$2\dfrac{dy}{dx}$$ (which is non-zero):

$$(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = 0$$

This is the required result.

Let $$y = x^2 + 3x + 2$$.

First derivative: $$\dfrac{dy}{dx} = 2x + 3$$.

Second derivative: $$\dfrac{d^2 y}{dx^2} = 2$$.

Let $$y = x^{20}$$.

First derivative: $$\dfrac{dy}{dx} = 20x^{19}$$.

Second derivative: $$\dfrac{d^2 y}{dx^2} = 20\cdot 19\,x^{18} = 380x^{18}$$.

Let $$y = x\cos x$$.

First derivative (product rule):

$$\dfrac{dy}{dx} = 1\cdot\cos x + x(-\sin x) = \cos x - x\sin x$$

Second derivative:

$$\dfrac{d^2 y}{dx^2} = -\sin x - (\sin x + x\cos x) = -2\sin x - x\cos x$$

Let $$y = \log x$$.

First derivative: $$\dfrac{dy}{dx} = \dfrac{1}{x} = x^{-1}$$.

Second derivative: $$\dfrac{d^2 y}{dx^2} = -x^{-2} = -\dfrac{1}{x^2}$$.

Let $$y = x^3\log x$$.

First derivative (product rule):

$$\dfrac{dy}{dx} = 3x^2\log x + x^3\cdot\dfrac{1}{x} = 3x^2\log x + x^2$$

Second derivative:

$$\dfrac{d^2 y}{dx^2} = \left(6x\log x + 3x^2\cdot\dfrac{1}{x}\right) + 2x = 6x\log x + 3x + 2x = 6x\log x + 5x$$

Let $$y = e^x\sin 5x$$.

First derivative (product rule):

$$\dfrac{dy}{dx} = e^x\sin 5x + e^x\cdot 5\cos 5x = e^x(\sin 5x + 5\cos 5x)$$

Second derivative:

$$\dfrac{d^2 y}{dx^2} = e^x(\sin 5x + 5\cos 5x) + e^x(5\cos 5x - 25\sin 5x)$$

$$= e^x(\sin 5x + 5\cos 5x + 5\cos 5x - 25\sin 5x) = e^x(10\cos 5x - 24\sin 5x)$$

Let $$y = e^{6x}\cos 3x$$.

First derivative (product rule):

$$\dfrac{dy}{dx} = 6e^{6x}\cos 3x + e^{6x}(-3\sin 3x) = e^{6x}(6\cos 3x - 3\sin 3x)$$

Second derivative:

$$\dfrac{d^2 y}{dx^2} = 6e^{6x}(6\cos 3x - 3\sin 3x) + e^{6x}(-18\sin 3x - 9\cos 3x)$$

$$= e^{6x}(36\cos 3x - 18\sin 3x - 18\sin 3x - 9\cos 3x) = e^{6x}(27\cos 3x - 36\sin 3x)$$

Let $$y = \tan^{-1} x$$.

First derivative: $$\dfrac{dy}{dx} = \dfrac{1}{1 + x^2} = (1 + x^2)^{-1}$$.

Second derivative (chain rule):

$$\dfrac{d^2 y}{dx^2} = -(1 + x^2)^{-2}\cdot 2x = \dfrac{-2x}{(1 + x^2)^2}$$

Let $$y = \log(\log x)$$.

First derivative: $$\dfrac{dy}{dx} = \dfrac{1}{\log x}\cdot\dfrac{1}{x} = \dfrac{1}{x\log x} = (x\log x)^{-1}$$.

Second derivative (chain rule), with $$\dfrac{d}{dx}(x\log x) = \log x + 1$$:

$$\dfrac{d^2 y}{dx^2} = -(x\log x)^{-2}\cdot(\log x + 1) = \dfrac{-(1 + \log x)}{(x\log x)^2}$$

Let $$y = \sin(\log x)$$.

First derivative: $$\dfrac{dy}{dx} = \cos(\log x)\cdot\dfrac{1}{x} = \dfrac{\cos(\log x)}{x}$$.

Second derivative (quotient rule):

$$\dfrac{d^2 y}{dx^2} = \dfrac{\left[-\sin(\log x)\cdot\frac{1}{x}\right]\cdot x - \cos(\log x)\cdot 1}{x^2}$$

$$= \dfrac{-\sin(\log x) - \cos(\log x)}{x^2}$$

Given $$y = 5\cos x - 3\sin x$$.

$$\dfrac{dy}{dx} = -5\sin x - 3\cos x$$

$$\dfrac{d^2 y}{dx^2} = -5\cos x + 3\sin x = -(5\cos x - 3\sin x) = -y$$

Therefore $$\dfrac{d^2 y}{dx^2} + y = 0$$.

Given $$y = \cos^{-1} x$$, so that $$x = \cos y$$.

First derivative:

$$\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 - x^2}} = -(1 - x^2)^{-1/2}$$

Second derivative (chain rule):

$$\dfrac{d^2 y}{dx^2} = -\left(-\dfrac{1}{2}\right)(1 - x^2)^{-3/2}\cdot(-2x) = \dfrac{-x}{(1 - x^2)^{3/2}}$$

Now express this in terms of $$y$$. Since $$x = \cos y$$ and $$1 - x^2 = 1 - \cos^2 y = \sin^2 y$$ (with $$\sin y \ge 0$$ for $$y \in [0, \pi]$$), we have $$(1 - x^2)^{3/2} = \sin^3 y$$. Hence

$$\dfrac{d^2 y}{dx^2} = \dfrac{-\cos y}{\sin^3 y}$$

Given $$y = 3\cos(\log x) + 4\sin(\log x)$$.

First derivative:

$$y_1 = -3\sin(\log x)\cdot\dfrac{1}{x} + 4\cos(\log x)\cdot\dfrac{1}{x} = \dfrac{4\cos(\log x) - 3\sin(\log x)}{x}$$

So $$x y_1 = 4\cos(\log x) - 3\sin(\log x)$$.

Differentiate this with respect to $$x$$ (product rule on the left side):

$$y_1 + x y_2 = -4\sin(\log x)\cdot\dfrac{1}{x} - 3\cos(\log x)\cdot\dfrac{1}{x} = \dfrac{-4\sin(\log x) - 3\cos(\log x)}{x}$$

Multiply both sides by $$x$$:

$$x y_1 + x^2 y_2 = -4\sin(\log x) - 3\cos(\log x) = -[3\cos(\log x) + 4\sin(\log x)] = -y$$

Therefore $$x^2 y_2 + x y_1 + y = 0$$.

Given $$y = A e^{mx} + B e^{nx}$$.

$$\dfrac{dy}{dx} = Am\,e^{mx} + Bn\,e^{nx}$$

$$\dfrac{d^2 y}{dx^2} = Am^2 e^{mx} + Bn^2 e^{nx}$$

Substitute into the left side of the given equation:

$$\dfrac{d^2 y}{dx^2} - (m + n)\dfrac{dy}{dx} + mn\,y$$

$$= (Am^2 e^{mx} + Bn^2 e^{nx}) - (m + n)(Am\,e^{mx} + Bn\,e^{nx}) + mn(A e^{mx} + B e^{nx})$$

Coefficient of $$e^{mx}$$: $$A\big[m^2 - (m + n)m + mn\big] = A[m^2 - m^2 - mn + mn] = 0$$.

Coefficient of $$e^{nx}$$: $$B\big[n^2 - (m + n)n + mn\big] = B[n^2 - mn - n^2 + mn] = 0$$.

Hence the whole expression equals $$0$$, proving $$\dfrac{d^2 y}{dx^2} - (m + n)\dfrac{dy}{dx} + mn\,y = 0$$.

Given $$y = 500e^{7x} + 600e^{-7x}$$.

$$\dfrac{dy}{dx} = 500\cdot 7\,e^{7x} + 600\cdot(-7)e^{-7x} = 3500e^{7x} - 4200e^{-7x}$$

$$\dfrac{d^2 y}{dx^2} = 3500\cdot 7\,e^{7x} - 4200\cdot(-7)e^{-7x} = 24500e^{7x} + 29400e^{-7x}$$

Factor out $$49$$:

$$\dfrac{d^2 y}{dx^2} = 49(500e^{7x} + 600e^{-7x}) = 49y$$

Hence $$\dfrac{d^2 y}{dx^2} = 49y$$.

Given $$e^y(x + 1) = 1$$, so $$e^y = \dfrac{1}{x + 1}$$.

Differentiate $$e^y(x + 1) = 1$$ with respect to $$x$$ (product rule):

$$e^y\dfrac{dy}{dx}(x + 1) + e^y = 0 \quad\Rightarrow\quad e^y\left[(x + 1)\dfrac{dy}{dx} + 1\right] = 0$$

Since $$e^y \ne 0$$, we get $$(x + 1)\dfrac{dy}{dx} + 1 = 0$$, hence

$$\dfrac{dy}{dx} = \dfrac{-1}{x + 1}$$

Differentiate again:

$$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\big[-(x + 1)^{-1}\big] = (x + 1)^{-2} = \dfrac{1}{(x + 1)^2}$$

Also $$\left(\dfrac{dy}{dx}\right)^2 = \left(\dfrac{-1}{x + 1}\right)^2 = \dfrac{1}{(x + 1)^2}$$.

Therefore $$\dfrac{d^2 y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2$$.

Given $$y = (\tan^{-1} x)^2$$.

First derivative (chain rule):

$$y_1 = 2\tan^{-1} x\cdot\dfrac{1}{1 + x^2}$$

Multiply both sides by $$(1 + x^2)$$:

$$(1 + x^2)y_1 = 2\tan^{-1} x$$

Differentiate both sides with respect to $$x$$ (product rule on the left):

$$(1 + x^2)y_2 + 2x\,y_1 = 2\cdot\dfrac{1}{1 + x^2}$$

Multiply both sides by $$(1 + x^2)$$:

$$(1 + x^2)^2 y_2 + 2x(1 + x^2)y_1 = 2$$

Since $$x^2 + 1 = 1 + x^2$$, this is exactly $$(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$$, as required.

Let $$y = f(x) = (\sin x)^{\sin x}$$. Since $$0 \lt x \lt \pi$$, we have $$\sin x \gt 0$$, so the expression is well defined and we may use logarithmic differentiation.

$$\log y = \sin x\cdot\log(\sin x)$$

Differentiate both sides with respect to $$x$$, using the product rule on the right:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \cos x\cdot\log(\sin x) + \sin x\cdot\dfrac{\cos x}{\sin x}$$

$$\dfrac{1}{y}\dfrac{dy}{dx} = \cos x\,\log(\sin x) + \cos x = \cos x\,[1 + \log(\sin x)]$$

Therefore

$$f'(x) = (\sin x)^{\sin x}\,\cos x\,[1 + \log(\sin x)]$$

Here both $$y$$ and $$x$$ are given in terms of the parameter $$t$$.

Differentiate $$y = a^{t + \frac{1}{t}}$$: taking logarithms, $$\log y = \left(t + \dfrac{1}{t}\right)\log a$$, so

$$\dfrac{1}{y}\dfrac{dy}{dt} = \left(1 - \dfrac{1}{t^2}\right)\log a \quad\Rightarrow\quad \dfrac{dy}{dt} = a^{t + \frac{1}{t}}\left(1 - \dfrac{1}{t^2}\right)\log a$$

Differentiate $$x = \left(t + \dfrac{1}{t}\right)^a$$: by the power rule (the exponent $$a$$ is constant),

$$\dfrac{dx}{dt} = a\left(t + \dfrac{1}{t}\right)^{a-1}\left(1 - \dfrac{1}{t^2}\right)$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{a^{t + \frac{1}{t}}\left(1 - \frac{1}{t^2}\right)\log a}{a\left(t + \frac{1}{t}\right)^{a-1}\left(1 - \frac{1}{t^2}\right)} = \dfrac{a^{t + \frac{1}{t}}\,\log a}{a\left(t + \frac{1}{t}\right)^{a-1}}$$

Let $$u = \sin^2 x$$ and $$v = e^{\cos x}$$. We are required to find $$\dfrac{du}{dv}$$.

Differentiate each with respect to $$x$$:

$$\dfrac{du}{dx} = 2\sin x\cos x = \sin 2x$$

$$\dfrac{dv}{dx} = e^{\cos x}\cdot(-\sin x) = -\sin x\,e^{\cos x}$$

Then

$$\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{2\sin x\cos x}{-\sin x\,e^{\cos x}} = \dfrac{-2\cos x}{e^{\cos x}}$$

Let $$y = (3x^2 - 9x + 5)^9$$. By the chain rule, with inner function $$u = 3x^2 - 9x + 5$$,

$$\dfrac{dy}{dx} = 9(3x^2 - 9x + 5)^8\cdot\dfrac{d}{dx}(3x^2 - 9x + 5)$$

$$= 9(3x^2 - 9x + 5)^8(6x - 9) = 27(2x - 3)(3x^2 - 9x + 5)^8$$

Let $$y = \sin^3 x + \cos^6 x$$. Differentiate each term by the chain rule.

$$\dfrac{d}{dx}\sin^3 x = 3\sin^2 x\cdot\cos x$$

$$\dfrac{d}{dx}\cos^6 x = 6\cos^5 x\cdot(-\sin x) = -6\sin x\cos^5 x$$

Therefore

$$\dfrac{dy}{dx} = 3\sin^2 x\cos x - 6\sin x\cos^5 x = 3\sin x\cos x\,(\sin x - 2\cos^4 x)$$

Let $$y = (5x)^{3\cos 2x}$$. Since the exponent involves $$x$$, use logarithmic differentiation.

$$\log y = 3\cos 2x\cdot\log(5x)$$

Differentiate both sides with respect to $$x$$, using the product rule on the right (with $$\dfrac{d}{dx}\log(5x) = \dfrac{1}{x}$$):

$$\dfrac{1}{y}\dfrac{dy}{dx} = 3(-2\sin 2x)\log(5x) + 3\cos 2x\cdot\dfrac{1}{x}$$

$$\dfrac{1}{y}\dfrac{dy}{dx} = -6\sin 2x\,\log(5x) + \dfrac{3\cos 2x}{x}$$

Therefore

$$\dfrac{dy}{dx} = (5x)^{3\cos 2x}\left[\dfrac{3\cos 2x}{x} - 6\sin 2x\,\log(5x)\right]$$

Let $$y = \sin^{-1}(x\sqrt{x}) = \sin^{-1}(x^{3/2})$$, $$0 \le x \le 1$$.

By the chain rule, with inner function $$u = x^{3/2}$$,

$$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - (x^{3/2})^2}}\cdot\dfrac{d}{dx}(x^{3/2}) = \dfrac{1}{\sqrt{1 - x^3}}\cdot\dfrac{3}{2}x^{1/2}$$

$$\dfrac{dy}{dx} = \dfrac{3\sqrt{x}}{2\sqrt{1 - x^3}}$$

Let $$y = \dfrac{\cos^{-1}(x/2)}{\sqrt{2x + 7}}$$. Apply the quotient rule with $$u = \cos^{-1}\dfrac{x}{2}$$ and $$v = (2x + 7)^{1/2}$$.

$$u' = \dfrac{-1}{\sqrt{1 - (x/2)^2}}\cdot\dfrac{1}{2} = \dfrac{-1}{2\sqrt{1 - x^2/4}} = \dfrac{-1}{\sqrt{4 - x^2}}$$

$$v' = \dfrac{1}{2}(2x + 7)^{-1/2}\cdot 2 = \dfrac{1}{\sqrt{2x + 7}}$$

By the quotient rule, $$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2}$$, with $$v^2 = 2x + 7$$:

$$\dfrac{dy}{dx} = \dfrac{\dfrac{-\sqrt{2x + 7}}{\sqrt{4 - x^2}} - \dfrac{\cos^{-1}(x/2)}{\sqrt{2x + 7}}}{2x + 7}$$

$$\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{4 - x^2}\,\sqrt{2x + 7}} - \dfrac{\cos^{-1}(x/2)}{(2x + 7)^{3/2}}$$

Let $$y = \cot^{-1}\left[\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right]$$, $$0 \lt x \lt \dfrac{\pi}{2}$$.

Use the identities $$1 + \sin x = \left(\cos\dfrac{x}{2} + \sin\dfrac{x}{2}\right)^2$$ and $$1 - \sin x = \left(\cos\dfrac{x}{2} - \sin\dfrac{x}{2}\right)^2$$.

For $$0 \lt x \lt \dfrac{\pi}{2}$$ we have $$\dfrac{x}{2} \in \left(0,\ \dfrac{\pi}{4}\right)$$, so $$\cos\dfrac{x}{2} \gt \sin\dfrac{x}{2} \gt 0$$. Taking positive square roots,

$$\sqrt{1 + \sin x} = \cos\dfrac{x}{2} + \sin\dfrac{x}{2}, \qquad \sqrt{1 - \sin x} = \cos\dfrac{x}{2} - \sin\dfrac{x}{2}$$

Then the numerator is $$2\cos\dfrac{x}{2}$$ and the denominator is $$2\sin\dfrac{x}{2}$$, so the bracket equals $$\cot\dfrac{x}{2}$$.

Hence $$y = \cot^{-1}\left(\cot\dfrac{x}{2}\right) = \dfrac{x}{2}$$, since $$\dfrac{x}{2} \in \left(0,\ \dfrac{\pi}{4}\right)$$ lies in the principal range $$(0,\ \pi)$$ of $$\cot^{-1}$$.

Differentiating,

$$\dfrac{dy}{dx} = \dfrac{1}{2}$$

Let $$y = (\log x)^{\log x}$$, $$x > 1$$. Use logarithmic differentiation.

$$\log y = \log x\cdot\log(\log x)$$

Differentiate both sides with respect to $$x$$, using the product rule on the right:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{x}\cdot\log(\log x) + \log x\cdot\dfrac{1}{\log x}\cdot\dfrac{1}{x}$$

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{\log(\log x)}{x} + \dfrac{1}{x} = \dfrac{1}{x}\big[1 + \log(\log x)\big]$$

Therefore

$$\dfrac{dy}{dx} = (\log x)^{\log x}\cdot\dfrac{1}{x}\big[1 + \log(\log x)\big]$$

Let $$y = \cos(a\cos x + b\sin x)$$. By the chain rule, with inner function $$u = a\cos x + b\sin x$$,

$$\dfrac{dy}{dx} = -\sin(a\cos x + b\sin x)\cdot\dfrac{d}{dx}(a\cos x + b\sin x)$$

Now $$\dfrac{d}{dx}(a\cos x + b\sin x) = -a\sin x + b\cos x$$. Therefore

$$\dfrac{dy}{dx} = -(b\cos x - a\sin x)\sin(a\cos x + b\sin x) = (a\sin x - b\cos x)\sin(a\cos x + b\sin x)$$

Let $$y = (\sin x - \cos x)^{(\sin x - \cos x)}$$, $$\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$$ (where $$\sin x - \cos x > 0$$). Use logarithmic differentiation.

$$\log y = (\sin x - \cos x)\cdot\log(\sin x - \cos x)$$

Let $$t = \sin x - \cos x$$, so $$\dfrac{dt}{dx} = \cos x + \sin x$$. Differentiate both sides with respect to $$x$$, using the product rule:

$$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{dt}{dx}\cdot\log t + t\cdot\dfrac{1}{t}\dfrac{dt}{dx} = \dfrac{dt}{dx}\big[\log t + 1\big]$$

Therefore

$$\dfrac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)}(\sin x + \cos x)\big[1 + \log(\sin x - \cos x)\big]$$

Let $$y = x^x + x^a + a^x + a^a$$, with $$a > 0$$ fixed and $$x > 0$$. Differentiate term by term.

$$x^x$$: let $$u = x^x$$, so $$\log u = x\log x$$ and $$\dfrac{1}{u}\dfrac{du}{dx} = \log x + 1$$, giving $$\dfrac{du}{dx} = x^x(1 + \log x)$$.

$$x^a$$: a power function, $$\dfrac{d}{dx}x^a = a x^{a-1}$$.

$$a^x$$: an exponential function, $$\dfrac{d}{dx}a^x = a^x\log a$$.

$$a^a$$: a constant, so its derivative is $$0$$.

Adding,

$$\dfrac{dy}{dx} = x^x(1 + \log x) + a x^{a-1} + a^x\log a$$

Let $$y = u + v$$, where $$u = x^{x^2 - 3}$$ and $$v = (x - 3)^{x^2}$$, for $$x > 3$$.

$$u = x^{x^2 - 3}$$: $$\log u = (x^2 - 3)\log x$$.

$$\dfrac{1}{u}\dfrac{du}{dx} = 2x\log x + (x^2 - 3)\cdot\dfrac{1}{x}$$

$$\dfrac{du}{dx} = x^{x^2 - 3}\left[2x\log x + \dfrac{x^2 - 3}{x}\right]$$

$$v = (x - 3)^{x^2}$$: $$\log v = x^2\log(x - 3)$$.

$$\dfrac{1}{v}\dfrac{dv}{dx} = 2x\log(x - 3) + x^2\cdot\dfrac{1}{x - 3}$$

$$\dfrac{dv}{dx} = (x - 3)^{x^2}\left[2x\log(x - 3) + \dfrac{x^2}{x - 3}\right]$$

Therefore

$$\dfrac{dy}{dx} = x^{x^2 - 3}\left[2x\log x + \dfrac{x^2 - 3}{x}\right] + (x - 3)^{x^2}\left[2x\log(x - 3) + \dfrac{x^2}{x - 3}\right]$$

Differentiate $$y$$ and $$x$$ with respect to the parameter $$t$$:

$$\dfrac{dy}{dt} = 12\sin t, \qquad \dfrac{dx}{dt} = 10(1 - \cos t)$$

Then

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{12\sin t}{10(1 - \cos t)} = \dfrac{6}{5}\cdot\dfrac{\sin t}{1 - \cos t}$$

Using $$\sin t = 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}$$ and $$1 - \cos t = 2\sin^2\dfrac{t}{2}$$,

$$\dfrac{\sin t}{1 - \cos t} = \dfrac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}} = \cot\dfrac{t}{2}$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{6}{5}\cot\dfrac{t}{2}$$

Let $$y = \sin^{-1} x + \sin^{-1}\sqrt{1 - x^2}$$, $$0 < x < 1$$.

Put $$\sin^{-1} x = \theta$$, so $$x = \sin\theta$$ with $$\theta \in \left(0, \dfrac{\pi}{2}\right)$$. Then

$$\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta = \sin\left(\dfrac{\pi}{2} - \theta\right)$$

Since $$\dfrac{\pi}{2} - \theta \in \left(0, \dfrac{\pi}{2}\right) \subset \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, we get $$\sin^{-1}\sqrt{1 - x^2} = \dfrac{\pi}{2} - \theta$$.

Hence

$$y = \theta + \left(\dfrac{\pi}{2} - \theta\right) = \dfrac{\pi}{2}$$

This is a constant, so

$$\dfrac{dy}{dx} = 0$$

Given $$x\sqrt{1 + y} + y\sqrt{1 + x} = 0$$, with $$-1 < x < 1$$ (and $$x \ne y$$).

Rewrite as $$x\sqrt{1 + y} = -y\sqrt{1 + x}$$ and square both sides:

$$x^2(1 + y) = y^2(1 + x)$$

$$x^2 + x^2 y = y^2 + y^2 x$$

$$x^2 - y^2 = y^2 x - x^2 y = xy(y - x)$$

$$(x - y)(x + y) = -xy(x - y)$$

Since $$x \ne y$$, divide both sides by $$(x - y)$$:

$$x + y = -xy \quad\Rightarrow\quad x + y + xy = 0 \quad\Rightarrow\quad y(1 + x) = -x$$

$$y = \dfrac{-x}{1 + x}$$

Differentiate by the quotient rule:

$$\dfrac{dy}{dx} = -\dfrac{(1)(1 + x) - x(1)}{(1 + x)^2} = -\dfrac{1 + x - x}{(1 + x)^2} = -\dfrac{1}{(1 + x)^2}$$

Hence $$\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}$$, as required.

Given $$(x - a)^2 + (y - b)^2 = c^2$$. Differentiate with respect to $$x$$:

$$2(x - a) + 2(y - b)\dfrac{dy}{dx} = 0 \quad\Rightarrow\quad (x - a) + (y - b)\dfrac{dy}{dx} = 0 \qquad \text{...(1)}$$

Differentiate (1) again with respect to $$x$$ (product rule on the second term):

$$1 + \left(\dfrac{dy}{dx}\right)^2 + (y - b)\dfrac{d^2 y}{dx^2} = 0$$

$$\Rightarrow\quad (y - b) = \dfrac{-\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]}{\dfrac{d^2 y}{dx^2}}$$

From (1), $$(x - a) = -(y - b)\dfrac{dy}{dx} = \dfrac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]\dfrac{dy}{dx}}{\dfrac{d^2 y}{dx^2}}$$.

Substitute both into $$(x - a)^2 + (y - b)^2 = c^2$$:

$$\dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2 y}{dx^2}\right)^2} + \dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2 y}{dx^2}\right)^2} = c^2$$

Take out the common factor in the numerator:

$$\dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2\left[\left(\frac{dy}{dx}\right)^2 + 1\right]}{\left(\frac{d^2 y}{dx^2}\right)^2} = c^2 \quad\Rightarrow\quad \dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^3}{\left(\frac{d^2 y}{dx^2}\right)^2} = c^2$$

Taking square roots,

$$\dfrac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2 y}{dx^2}} = \pm c$$

This is a constant ($$\pm c$$), independent of $$a$$ and $$b$$. Hence proved.

Given $$\cos y = x\cos(a + y)$$. Solving for $$x$$,

$$x = \dfrac{\cos y}{\cos(a + y)}$$

Differentiate $$x$$ with respect to $$y$$ using the quotient rule:

$$\dfrac{dx}{dy} = \dfrac{(-\sin y)\cos(a + y) - \cos y\,\big(-\sin(a + y)\big)}{\cos^2(a + y)}$$

$$= \dfrac{\cos y\sin(a + y) - \sin y\cos(a + y)}{\cos^2(a + y)}$$

The numerator is $$\sin\big((a + y) - y\big) = \sin a$$. Hence

$$\dfrac{dx}{dy} = \dfrac{\sin a}{\cos^2(a + y)}$$

Therefore

$$\dfrac{dy}{dx} = \dfrac{1}{dx/dy} = \dfrac{\cos^2(a + y)}{\sin a}$$

Hence proved. (The condition $$\cos a \ne \pm 1$$ ensures $$\sin a \ne 0$$.)

Differentiate $$x$$ and $$y$$ with respect to $$t$$, using the product rule on $$t\sin t$$ and $$t\cos t$$:

$$\dfrac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = a t\cos t$$

$$\dfrac{dy}{dt} = a(\cos t - \cos t + t\sin t) = a t\sin t$$

Hence

$$\dfrac{dy}{dx} = \dfrac{a t\sin t}{a t\cos t} = \tan t$$

Differentiate $$\dfrac{dy}{dx} = \tan t$$ with respect to $$x$$, again via the parameter $$t$$:

$$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(\tan t) = \dfrac{d}{dt}(\tan t)\cdot\dfrac{dt}{dx} = \sec^2 t\cdot\dfrac{1}{dx/dt}$$

$$= \sec^2 t\cdot\dfrac{1}{a t\cos t} = \dfrac{\sec^2 t}{a t\cos t} = \dfrac{\sec^3 t}{a t}$$

Write $$f(x) = |x|^3$$ as a piecewise function. For $$x \ge 0$$, $$|x| = x$$, so $$f(x) = x^3$$; for $$x \lt 0$$, $$|x| = -x$$, so $$f(x) = (-x)^3 = -x^3$$. Thus

$$f(x) = \begin{cases} x^3, & \text{if } x \ge 0 \\ -x^3, & \text{if } x \lt 0 \end{cases}$$

First derivative. For $$x \gt 0$$, $$f'(x) = 3x^2$$; for $$x \lt 0$$, $$f'(x) = -3x^2$$. At $$x = 0$$, the difference quotient is $$\dfrac{f(0 + h) - f(0)}{h} = \dfrac{f(h)}{h}$$; for $$h \gt 0$$ this equals $$\dfrac{h^3}{h} = h^2$$ and for $$h \lt 0$$ it equals $$\dfrac{-h^3}{h} = -h^2$$, and both tend to $$0$$ as $$h \to 0$$. Hence $$f'(0) = 0$$. Therefore

$$f'(x) = \begin{cases} 3x^2, & \text{if } x \ge 0 \\ -3x^2, & \text{if } x \lt 0 \end{cases}$$

so $$f'$$ is defined for every real $$x$$ (in particular $$f'(0) = 0$$).

Second derivative. For $$x \gt 0$$, $$f''(x) = 6x$$; for $$x \lt 0$$, $$f''(x) = -6x$$. At $$x = 0$$, the difference quotient of $$f'$$ is $$\dfrac{f'(0 + h) - f'(0)}{h} = \dfrac{f'(h)}{h}$$; for $$h \gt 0$$ this equals $$\dfrac{3h^2}{h} = 3h \to 0$$ and for $$h \lt 0$$ it equals $$\dfrac{-3h^2}{h} = -3h \to 0$$. Hence $$f''(0) = 0$$.

Therefore $$f''(x)$$ exists for every real $$x$$, and

$$f''(x) = \begin{cases} 6x, & \text{if } x \ge 0 \\ -6x, & \text{if } x \lt 0 \end{cases} = 6|x|$$

We are given the identity

$$\sin(A + B) = \sin A\cos B + \cos A\sin B$$

Treat $$A$$ as a variable and $$B$$ as a constant, and differentiate both sides with respect to $$A$$.

Left side: by the chain rule, $$\dfrac{d}{dA}\sin(A + B) = \cos(A + B)\cdot\dfrac{d}{dA}(A + B) = \cos(A + B)$$.

Right side: since $$\cos B$$ and $$\sin B$$ are constants,

$$\dfrac{d}{dA}\big[\sin A\cos B + \cos A\sin B\big] = \cos A\cos B - \sin A\sin B$$

Equating the two results,

$$\cos(A + B) = \cos A\cos B - \sin A\sin B$$

This is the required sum formula for cosines.

Yes, such a function exists. Consider

$$f(x) = |x| + |x - 1|$$

Continuity: The modulus function is continuous everywhere, so both $$|x|$$ and $$|x - 1|$$ are continuous everywhere. Their sum $$f$$ is therefore continuous at every real number.

Differentiability: The function $$|x|$$ fails to be differentiable only at $$x = 0$$ (it has a corner there), and $$|x - 1|$$ fails to be differentiable only at $$x = 1$$.

• At $$x = 0$$: $$|x - 1|$$ is differentiable but $$|x|$$ is not, so the sum $$f$$ is not differentiable.
• At $$x = 1$$: $$|x|$$ is differentiable but $$|x - 1|$$ is not, so the sum $$f$$ is not differentiable.
• At every other point both terms are differentiable, so $$f$$ is differentiable.

Hence $$f(x) = |x| + |x - 1|$$ is continuous everywhere but not differentiable at exactly the two points $$x = 0$$ and $$x = 1$$. So yes, such a function does exist.

Given $$y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$$, where $$l, m, n, a, b, c$$ are constants.

Expand the determinant along the first row:

$$y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)$$

Here $$(mc - nb)$$, $$(lc - na)$$ and $$(lb - ma)$$ are all constants. Differentiate $$y$$ with respect to $$x$$:

$$\dfrac{dy}{dx} = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)$$

But the right-hand side is precisely the first-row expansion of the determinant

$$\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$$

Therefore

$$\dfrac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$$

Hence proved.

Given $$y = e^{a\cos^{-1} x}$$, $$-1 \le x \le 1$$.

Differentiate with respect to $$x$$, using the chain rule:

$$\dfrac{dy}{dx} = e^{a\cos^{-1} x}\cdot a\cdot\dfrac{-1}{\sqrt{1 - x^2}} = \dfrac{-a\,y}{\sqrt{1 - x^2}}$$

So $$\sqrt{1 - x^2}\,\dfrac{dy}{dx} = -a\,y$$. Square both sides:

$$(1 - x^2)\left(\dfrac{dy}{dx}\right)^2 = a^2 y^2$$

Differentiate both sides with respect to $$x$$:

$$(1 - x^2)\cdot 2\dfrac{dy}{dx}\dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2(-2x) = a^2\cdot 2y\dfrac{dy}{dx}$$

Divide throughout by $$2\dfrac{dy}{dx}$$ (which is non-zero):

$$(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = a^2 y$$

Therefore

$$(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} - a^2 y = 0$$

Hence proved.