Chapter 12: Limits and Derivatives

The expression $$x + 3$$ is a polynomial, so its limit is found by directly substituting $$x = 3$$.

$$\lim_{x \to 3}(x + 3) = 3 + 3 = 6$$

The expression $$x - \dfrac{22}{7}$$ is a polynomial in $$x$$ (the term $$\dfrac{22}{7}$$ is a constant). Its limit is found by directly substituting $$x = \pi$$.

$$\lim_{x \to \pi}\left(x - \frac{22}{7}\right) = \pi - \frac{22}{7}$$

Here $$\pi$$ is a constant, and $$\pi r^2$$ is a polynomial in the variable $$r$$. Its limit is found by directly substituting $$r = 1$$.

$$\lim_{r \to 1}\pi r^2 = \pi\,(1)^2 = \pi$$

Numerator and denominator are polynomials. The denominator at $$x = 4$$ is $$4 - 2 = 2 \neq 0$$, so the limit of the quotient is the quotient of the limits, found by direct substitution.

$$\lim_{x \to 4}\frac{4x + 3}{x - 2} = \frac{4(4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2}$$

The denominator at $$x = -1$$ is $$-1 - 1 = -2 \neq 0$$, so the limit may be found by direct substitution.

Numerator at $$x = -1$$: $$(-1)^{10} + (-1)^5 + 1 = 1 + (-1) + 1 = 1$$.

$$\lim_{x \to -1}\frac{x^{10} + x^5 + 1}{x - 1} = \frac{1}{-2} = -\frac{1}{2}$$

At $$x = 0$$: numerator $$(0 + 1)^5 - 1 = 0$$ and denominator $$= 0$$, the form $$\frac{0}{0}$$. Expand $$(x + 1)^5$$ using the binomial theorem.

$$(x + 1)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$$

So

$$(x + 1)^5 - 1 = 5x + 10x^2 + 10x^3 + 5x^4 + x^5$$

Divide by $$x$$ (allowed since $$x \neq 0$$):

$$\frac{(x + 1)^5 - 1}{x} = 5 + 10x + 10x^2 + 5x^3 + x^4$$

Now substitute $$x = 0$$:

$$\lim_{x \to 0}\frac{(x + 1)^5 - 1}{x} = 5 + 0 + 0 + 0 + 0 = 5$$

At $$x = 2$$: numerator $$3(4) - 2 - 10 = 0$$ and denominator $$4 - 4 = 0$$, the form $$\frac{0}{0}$$. Factorise both.

Numerator: $$3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x - 2) + 5(x - 2) = (x - 2)(3x + 5)$$.

Denominator: $$x^2 - 4 = (x - 2)(x + 2)$$.

$$\lim_{x \to 2}\frac{(x - 2)(3x + 5)}{(x - 2)(x + 2)} = \lim_{x \to 2}\frac{3x + 5}{x + 2}$$

Substitute $$x = 2$$:

$$= \frac{3(2) + 5}{2 + 2} = \frac{11}{4}$$

At $$x = 3$$: numerator $$81 - 81 = 0$$ and denominator $$2(9) - 15 - 3 = 0$$, the form $$\frac{0}{0}$$. Factorise both.

Numerator: $$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$$.

Denominator: $$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (x - 3)(2x + 1)$$.

$$\lim_{x \to 3}\frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)} = \lim_{x \to 3}\frac{(x + 3)(x^2 + 9)}{2x + 1}$$

Substitute $$x = 3$$:

$$= \frac{(3 + 3)(9 + 9)}{2(3) + 1} = \frac{6 \times 18}{7} = \frac{108}{7}$$

The denominator at $$x = 0$$ is $$c(0) + 1 = 1 \neq 0$$, so the limit is found by direct substitution.

$$\lim_{x \to 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{b}{1} = b$$

At $$z = 1$$ both numerator and denominator vanish, giving $$\frac{0}{0}$$. Substitute $$z^{1/6} = t$$, so that $$z^{1/3} = \left(z^{1/6}\right)^2 = t^2$$.

As $$z \to 1$$, we have $$t = z^{1/6} \to 1$$. The limit becomes

$$\lim_{t \to 1}\frac{t^2 - 1}{t - 1} = \lim_{t \to 1}\frac{(t - 1)(t + 1)}{t - 1} = \lim_{t \to 1}(t + 1)$$

Substitute $$t = 1$$:

$$= 1 + 1 = 2$$

Substitute $$x = 1$$ in numerator and denominator.

Numerator at $$x = 1$$: $$a(1)^2 + b(1) + c = a + b + c$$.

Denominator at $$x = 1$$: $$c(1)^2 + b(1) + a = a + b + c$$.

Since it is given that $$a + b + c \neq 0$$, the denominator is non-zero, and the limit is

$$\lim_{x \to 1}\frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a + b + c}{a + b + c} = 1$$

At $$x = -2$$ the numerator $$\dfrac{1}{x} + \dfrac{1}{2} = -\dfrac{1}{2} + \dfrac{1}{2} = 0$$ and the denominator $$= 0$$, the form $$\frac{0}{0}$$. Simplify the numerator first.

$$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$$

So the whole expression is

$$\frac{\dfrac{x + 2}{2x}}{x + 2} = \frac{x + 2}{2x} \cdot \frac{1}{x + 2} = \frac{1}{2x}$$

The cancellation of $$x + 2$$ is valid since $$x \to -2$$ means $$x \neq -2$$. Now substitute $$x = -2$$:

$$\lim_{x \to -2}\frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$$

We use the standard limit $$\lim_{\theta \to 0}\dfrac{\sin \theta}{\theta} = 1$$. Rewrite the expression so that $$\sin ax$$ is divided by its argument $$ax$$.

$$\frac{\sin ax}{bx} = \frac{\sin ax}{ax} \cdot \frac{ax}{bx} = \frac{\sin ax}{ax} \cdot \frac{a}{b}$$

As $$x \to 0$$, $$ax \to 0$$, so $$\dfrac{\sin ax}{ax} \to 1$$. Therefore

$$\lim_{x \to 0}\frac{\sin ax}{bx} = 1 \cdot \frac{a}{b} = \frac{a}{b}$$

Rewrite so that each sine is divided by its own argument:

$$\frac{\sin ax}{\sin bx} = \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} = \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b}$$

As $$x \to 0$$, both $$ax \to 0$$ and $$bx \to 0$$, so

$$\lim_{x \to 0}\frac{\sin ax}{ax} = 1, \qquad \lim_{x \to 0}\frac{bx}{\sin bx} = 1$$

Therefore

$$\lim_{x \to 0}\frac{\sin ax}{\sin bx} = 1 \cdot 1 \cdot \frac{a}{b} = \frac{a}{b}$$

Put $$y = \pi - x$$. As $$x \to \pi$$, we have $$y \to 0$$. The expression becomes

$$\frac{\sin(\pi - x)}{\pi(\pi - x)} = \frac{\sin y}{\pi\,y} = \frac{1}{\pi} \cdot \frac{\sin y}{y}$$

As $$y \to 0$$, $$\dfrac{\sin y}{y} \to 1$$. Therefore

$$\lim_{x \to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)} = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}$$

The function $$\cos x$$ and the polynomial $$\pi - x$$ are both well-behaved at $$x = 0$$, and the denominator $$\pi - 0 = \pi \neq 0$$. So the limit is found by direct substitution.

$$\lim_{x \to 0}\frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi}$$

At $$x = 0$$ both numerator and denominator are $$0$$. Use the identity $$1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$$, which gives

$$\cos 2x - 1 = -2\sin^2 x, \qquad \cos x - 1 = -2\sin^2\frac{x}{2}$$

So

$$\frac{\cos 2x - 1}{\cos x - 1} = \frac{-2\sin^2 x}{-2\sin^2\frac{x}{2}} = \frac{\sin^2 x}{\sin^2\frac{x}{2}}$$

Rewrite to use $$\dfrac{\sin\theta}{\theta} \to 1$$:

$$\frac{\sin^2 x}{\sin^2\frac{x}{2}} = \left(\frac{\sin x}{x}\right)^2 \cdot \left(\frac{\frac{x}{2}}{\sin\frac{x}{2}}\right)^2 \cdot \frac{x^2}{\left(\frac{x}{2}\right)^2}$$

Here $$\dfrac{x^2}{\left(x/2\right)^2} = \dfrac{x^2}{x^2/4} = 4$$. As $$x \to 0$$, $$\dfrac{\sin x}{x} \to 1$$ and $$\dfrac{x/2}{\sin(x/2)} \to 1$$.

$$\lim_{x \to 0}\frac{\cos 2x - 1}{\cos x - 1} = (1)^2 \cdot (1)^2 \cdot 4 = 4$$

Take $$x$$ common in the numerator:

$$\frac{ax + x\cos x}{b\sin x} = \frac{x(a + \cos x)}{b\sin x} = \frac{1}{b} \cdot \frac{x}{\sin x} \cdot (a + \cos x)$$

As $$x \to 0$$: $$\dfrac{x}{\sin x} \to 1$$ and $$a + \cos x \to a + \cos 0 = a + 1$$.

$$\lim_{x \to 0}\frac{ax + x\cos x}{b\sin x} = \frac{1}{b} \cdot 1 \cdot (a + 1) = \frac{a + 1}{b}$$

Write $$\sec x = \dfrac{1}{\cos x}$$:

$$x\sec x = \frac{x}{\cos x}$$

At $$x = 0$$ the denominator $$\cos 0 = 1 \neq 0$$, so the limit is found by direct substitution.

$$\lim_{x \to 0} x\sec x = \frac{0}{\cos 0} = \frac{0}{1} = 0$$

Divide both numerator and denominator by $$x$$ (valid since $$x \neq 0$$):

$$\frac{\sin ax + bx}{ax + \sin bx} = \frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}}$$

Now $$\dfrac{\sin ax}{x} = a \cdot \dfrac{\sin ax}{ax}$$ and $$\dfrac{\sin bx}{x} = b \cdot \dfrac{\sin bx}{bx}$$.

As $$x \to 0$$: $$\dfrac{\sin ax}{ax} \to 1$$ and $$\dfrac{\sin bx}{bx} \to 1$$, so

$$\frac{\sin ax}{x} \to a, \qquad \frac{\sin bx}{x} \to b$$

Therefore

$$\lim_{x \to 0}\frac{\sin ax + bx}{ax + \sin bx} = \frac{a + b}{a + b} = 1$$

(The condition $$a + b \neq 0$$ ensures the denominator of this last fraction is non-zero.)

Write both functions in terms of $$\sin x$$ and $$\cos x$$:

$$\csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}$$

Use the half-angle identities $$1 - \cos x = 2\sin^2\dfrac{x}{2}$$ and $$\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$$:

$$\frac{1 - \cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2}$$

As $$x \to 0$$, $$\dfrac{x}{2} \to 0$$, so

$$\lim_{x \to 0}(\csc x - \cot x) = \lim_{x \to 0}\tan\frac{x}{2} = \tan 0 = 0$$

Put $$y = x - \dfrac{\pi}{2}$$, so that $$x = y + \dfrac{\pi}{2}$$. As $$x \to \dfrac{\pi}{2}$$, we have $$y \to 0$$.

Then $$2x = 2y + \pi$$, and since $$\tan$$ has period $$\pi$$,

$$\tan 2x = \tan(2y + \pi) = \tan 2y$$

The limit becomes

$$\lim_{y \to 0}\frac{\tan 2y}{y} = \lim_{y \to 0}\frac{\tan 2y}{2y} \cdot 2$$

Using $$\lim_{\theta \to 0}\dfrac{\tan\theta}{\theta} = 1$$ with $$\theta = 2y \to 0$$:

$$= 1 \cdot 2 = 2$$

Limit at $$x = 0$$. The function is given by different formulas on the two sides of $$0$$, so we compute the two one-sided limits separately.

Left-hand limit. For $$x \le 0$$ — in particular for $$x \lt 0$$ — the function is $$f(x) = 2x + 3$$. Therefore

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-}(2x + 3) = 2(0) + 3 = 3$$

Right-hand limit. For $$x \gt 0$$, the function is $$f(x) = 3(x + 1)$$. Therefore

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x + 1) = 3(0 + 1) = 3$$

The left-hand limit and the right-hand limit are both equal to $$3$$, so the limit exists and

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

Limit at $$x = 1$$. Every $$x$$ in a small neighbourhood of $$1$$ satisfies $$x \gt 0$$, so $$f(x) = 3(x + 1)$$ throughout that neighbourhood. The limit is found by direct substitution:

$$\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x + 1) = 3(1 + 1) = 6$$

The function changes formula at $$x = 1$$, so we compute the two one-sided limits.

For $$x \le 1$$, $$f(x) = x^2 - 1$$. Left-hand limit:

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-}(x^2 - 1) = (1)^2 - 1 = 0$$

For $$x > 1$$, $$f(x) = -x^2 - 1$$. Right-hand limit:

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

Since the left-hand limit $$(0)$$ is not equal to the right-hand limit $$(-2)$$, the limit does not exist.

$$\lim_{x \to 1} f(x) \text{ does not exist.}$$

Recall that $$|x| = x$$ when $$x \gt 0$$, and $$|x| = -x$$ when $$x \lt 0$$.

Right-hand limit. For $$x \gt 0$$, $$f(x) = \dfrac{|x|}{x} = \dfrac{x}{x} = 1$$. Hence

$$\lim_{x \to 0^+} f(x) = 1$$

Left-hand limit. For $$x \lt 0$$, $$f(x) = \dfrac{|x|}{x} = \dfrac{-x}{x} = -1$$. Hence

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

The right-hand limit $$(1)$$ and the left-hand limit $$(-1)$$ are unequal, so the limit does not exist.

$$\lim_{x \to 0} f(x) \text{ does not exist.}$$

Recall that $$|x| = x$$ when $$x \gt 0$$, and $$|x| = -x$$ when $$x \lt 0$$.

Right-hand limit. For $$x \gt 0$$, $$f(x) = \dfrac{x}{|x|} = \dfrac{x}{x} = 1$$. Hence

$$\lim_{x \to 0^+} f(x) = 1$$

Left-hand limit. For $$x \lt 0$$, $$f(x) = \dfrac{x}{|x|} = \dfrac{x}{-x} = -1$$. Hence

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

The right-hand limit $$(1)$$ and the left-hand limit $$(-1)$$ are unequal, so the limit does not exist.

$$\lim_{x \to 0} f(x) \text{ does not exist.}$$

For all values of $$x$$ near $$5$$ we have $$x > 0$$, so $$|x| = x$$ throughout that neighbourhood. Hence near $$x = 5$$,

$$f(x) = |x| - 5 = x - 5$$

This is a polynomial, so its limit is found by direct substitution:

$$\lim_{x \to 5} f(x) = \lim_{x \to 5}(x - 5) = 5 - 5 = 0$$

(Both one-sided limits equal $$0$$, confirming the limit exists.)

From the definition, $$f(1) = 4$$.

Compute the one-sided limits at $$x = 1$$.

For $$x < 1$$, $$f(x) = a + bx$$, so the left-hand limit is

$$\lim_{x \to 1^-} f(x) = a + b(1) = a + b$$

For $$x > 1$$, $$f(x) = b - ax$$, so the right-hand limit is

$$\lim_{x \to 1^+} f(x) = b - a(1) = b - a$$

For $$\lim_{x \to 1} f(x)$$ to exist, the two one-sided limits must be equal, and the condition $$\lim_{x \to 1} f(x) = f(1)$$ requires both to equal $$4$$:

$$a + b = 4 \qquad \text{and} \qquad b - a = 4$$

Adding the two equations: $$2b = 8$$, so $$b = 4$$. Subtracting the second from the first: $$2a = 0$$, so $$a = 0$$.

$$a = 0, \qquad b = 4$$

The function $$f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$$ is a polynomial, so its limit at any point is obtained by direct substitution.

Limit as $$x \to a_1$$. Substitute $$x = a_1$$. The first factor becomes $$a_1 - a_1 = 0$$, which makes the whole product $$0$$:

$$\lim_{x \to a_1} f(x) = (a_1 - a_1)(a_1 - a_2)\cdots(a_1 - a_n) = 0$$

Limit as $$x \to a$$, where $$a \neq a_1, a_2, \ldots, a_n$$. Substitute $$x = a$$:

$$\lim_{x \to a} f(x) = (a - a_1)(a - a_2)\cdots(a - a_n)$$

Since $$a$$ differs from every $$a_i$$, none of these factors is zero, so the value is a non-zero product.

First simplify $$f$$ on each side of $$0$$. For $$x \lt 0$$, $$|x| = -x$$, so $$f(x) = |x| + 1 = -x + 1$$. For $$x \gt 0$$, $$|x| = x$$, so $$f(x) = |x| - 1 = x - 1$$.

Case $$a \lt 0$$. Every point in a small neighbourhood of such an $$a$$ satisfies $$x \lt 0$$, so there $$f(x) = -x + 1$$, a polynomial. Hence

$$\lim_{x \to a} f(x) = -a + 1$$

and the limit exists.

Case $$a \gt 0$$. Every point in a small neighbourhood of such an $$a$$ satisfies $$x \gt 0$$, so there $$f(x) = x - 1$$, a polynomial. Hence

$$\lim_{x \to a} f(x) = a - 1$$

and the limit exists.

Case $$a = 0$$. Here we must examine the one-sided limits separately:

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

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

Since $$1 \neq -1$$, the left-hand and right-hand limits differ, so the limit does not exist at $$a = 0$$.

Therefore $$\lim_{x \to a} f(x)$$ exists for every real number $$a$$ except $$a = 0$$.

We are given that

$$\lim_{x \to 1}\frac{f(x) - 2}{x^2 - 1} = \pi$$

Write $$f(x) - 2$$ as a product so we can use the product rule for limits:

$$f(x) - 2 = \frac{f(x) - 2}{x^2 - 1} \cdot (x^2 - 1)$$

Taking limits as $$x \to 1$$, and using that the limit of a product is the product of the limits:

$$\lim_{x \to 1}\big(f(x) - 2\big) = \lim_{x \to 1}\frac{f(x) - 2}{x^2 - 1} \cdot \lim_{x \to 1}(x^2 - 1) = \pi \cdot (1^2 - 1) = \pi \cdot 0 = 0$$

Hence $$\lim_{x \to 1} f(x) - 2 = 0$$, which gives

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

Limit at $$x = 0$$. The formula for $$f$$ changes at $$x = 0$$, so we compute the two one-sided limits.

Left-hand limit. For $$x \lt 0$$, $$f(x) = mx^2 + n$$, so

$$\lim_{x \to 0^-} f(x) = m(0)^2 + n = n$$

Right-hand limit. For $$0 \le x \le 1$$, $$f(x) = nx + m$$, so

$$\lim_{x \to 0^+} f(x) = n(0) + m = m$$

For $$\lim_{x \to 0} f(x)$$ to exist, these one-sided limits must be equal:

$$n = m$$

Limit at $$x = 1$$. The formula for $$f$$ changes at $$x = 1$$, so we again compute the two one-sided limits.

Left-hand limit. For $$0 \le x \le 1$$, $$f(x) = nx + m$$, so

$$\lim_{x \to 1^-} f(x) = n(1) + m = n + m$$

Right-hand limit. For $$x \gt 1$$, $$f(x) = nx^3 + m$$, so

$$\lim_{x \to 1^+} f(x) = n(1)^3 + m = n + m$$

These two one-sided limits are equal for every choice of $$m$$ and $$n$$, so $$\lim_{x \to 1} f(x)$$ always exists.

Conclusion. The limit at $$x = 1$$ exists for all integers $$m$$ and $$n$$. The limit at $$x = 0$$ exists only when $$m = n$$. Hence both limits exist precisely when $$m = n$$ (for any integer value).

By the first principle, the derivative of $$f$$ at $$x = 2$$ is

$$f'(2) = \lim_{h \to 0}\frac{f(2 + h) - f(2)}{h}$$

Here $$f(x) = 3x$$, so $$f(2 + h) = 3(2 + h) = 6 + 3h$$ and $$f(2) = 3(2) = 6$$.

$$f'(2) = \lim_{h \to 0}\frac{(6 + 3h) - 6}{h} = \lim_{h \to 0}\frac{3h}{h} = \lim_{h \to 0} 3 = 3$$

By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$

Compute $$f(x + h)$$:

$$f(x + h) = 2(x + h)^2 + 3(x + h) - 5 = 2x^2 + 4xh + 2h^2 + 3x + 3h - 5$$

Subtract $$f(x) = 2x^2 + 3x - 5$$:

$$f(x + h) - f(x) = 4xh + 2h^2 + 3h = h(4x + 2h + 3)$$

Therefore

$$f'(x) = \lim_{h \to 0}\frac{h(4x + 2h + 3)}{h} = \lim_{h \to 0}(4x + 2h + 3) = 4x + 3$$

At $$x = -1$$:

$$f'(-1) = 4(-1) + 3 = -1$$

Proof of the identity. From $$f'(x) = 4x + 3$$ we get $$f'(0) = 4(0) + 3 = 3$$. Then

$$f'(0) + 3f'(-1) = 3 + 3(-1) = 3 - 3 = 0$$

Hence $$f'(0) + 3f'(-1) = 0$$, as required.

Let $$f(x) = \sin x$$. By the first principle, the derivative at $$x = 0$$ is

$$f'(0) = \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0}\frac{\sin(0 + h) - \sin 0}{h}$$

Since $$\sin 0 = 0$$, this becomes

$$f'(0) = \lim_{h \to 0}\frac{\sin h}{h} = 1$$

using the standard limit $$\lim_{h \to 0}\dfrac{\sin h}{h} = 1$$.

The function $$f(x) = 3$$ is constant: it takes the value $$3$$ for every $$x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{3 - 3}{h} = \lim_{h \to 0}\frac{0}{h} = \lim_{h \to 0} 0 = 0$$

This holds for every $$x$$. In particular,

$$f'(0) = 0 \qquad \text{and} \qquad f'(3) = 0$$

By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$

Here $$f(x + h) = 10(x + h) = 10x + 10h$$, so

$$f(x + h) - f(x) = (10x + 10h) - 10x = 10h$$

Therefore

$$f'(x) = \lim_{h \to 0}\frac{10h}{h} = \lim_{h \to 0} 10 = 10$$

By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$

Compute $$f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$$, so

$$f(x + h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2 = h(2x + h)$$

Therefore

$$f'(x) = \lim_{h \to 0}\frac{h(2x + h)}{h} = \lim_{h \to 0}(2x + h) = 2x$$

The function $$f(x) = a$$ has the same value $$a$$ for every $$x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{a - a}{h} = \lim_{h \to 0}\frac{0}{h} = \lim_{h \to 0} 0 = 0$$

So the derivative of a constant function is $$0$$ everywhere.

By the first principle,

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

Combine the fractions in the numerator:

$$\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}$$

So

$$f'(x) = \lim_{h \to 0}\frac{1}{h} \cdot \frac{-h}{x(x + h)} = \lim_{h \to 0}\frac{-1}{x(x + h)}$$

Now let $$h \to 0$$:

$$f'(x) = \frac{-1}{x \cdot x} = -\frac{1}{x^2}$$

Let $$f(x) = 6x^{100} - x^{55} + x$$. We use the power rule $$\dfrac{d}{dx}x^n = nx^{n-1}$$ together with the fact that the derivative of a sum is the sum of the derivatives.

$$\frac{d}{dx}\big(6x^{100}\big) = 6 \cdot 100 x^{99} = 600x^{99}$$

$$\frac{d}{dx}\big(x^{55}\big) = 55x^{54}$$

$$\frac{d}{dx}(x) = 1$$

Combining these,

$$f'(x) = 600x^{99} - 55x^{54} + 1$$

Differentiate term by term using the power rule $$\dfrac{d}{dx}x^n = nx^{n-1}$$. The constant term $$1$$ has derivative $$0$$.

$$f'(x) = 0 + 1 + 2x + 3x^2 + \ldots + 50x^{49}$$

Now evaluate at $$x = 1$$. Every power of $$1$$ equals $$1$$, so

$$f'(1) = 1 + 2 + 3 + \ldots + 50$$

This is the sum of the first $$50$$ natural numbers. Using $$1 + 2 + \ldots + n = \dfrac{n(n + 1)}{2}$$ with $$n = 50$$:

$$f'(1) = \frac{50 \times 51}{2} = \frac{2550}{2} = 1275$$

First simplify the function by splitting the fraction:

$$f(x) = \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + \frac{1}{x}$$

Now differentiate. The constant $$1$$ has derivative $$0$$, and (from Example 12) $$\dfrac{d}{dx}\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^2}$$.

$$f'(x) = 0 + \left(-\frac{1}{x^2}\right) = -\frac{1}{x^2}$$

Let $$f(x) = \sin x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{\sin(x + h) - \sin x}{h}$$

Use the difference-to-product identity $$\sin C - \sin D = 2\cos\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$$ with $$C = x + h$$ and $$D = x$$:

$$\sin(x + h) - \sin x = 2\cos\left(x + \frac{h}{2}\right)\sin\frac{h}{2}$$

So

$$f'(x) = \lim_{h \to 0}\frac{2\cos\left(x + \frac{h}{2}\right)\sin\frac{h}{2}}{h} = \lim_{h \to 0}\cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\frac{h}{2}}{\frac{h}{2}}$$

As $$h \to 0$$: $$\cos\left(x + \dfrac{h}{2}\right) \to \cos x$$ and $$\dfrac{\sin(h/2)}{h/2} \to 1$$. Therefore

$$\frac{d}{dx}(\sin x) = \cos x$$

Let $$f(x) = \tan x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{\tan(x + h) - \tan x}{h}$$

Write each tangent as $$\dfrac{\sin}{\cos}$$ and combine over a common denominator:

$$\tan(x + h) - \tan x = \frac{\sin(x + h)}{\cos(x + h)} - \frac{\sin x}{\cos x} = \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{\cos(x + h)\cos x}$$

The numerator is $$\sin\big((x + h) - x\big) = \sin h$$. So

$$\tan(x + h) - \tan x = \frac{\sin h}{\cos(x + h)\cos x}$$

Therefore

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

As $$h \to 0$$: $$\dfrac{\sin h}{h} \to 1$$ and $$\cos(x + h) \to \cos x$$, giving

$$f'(x) = 1 \cdot \frac{1}{\cos x \cos x} = \frac{1}{\cos^2 x} = \sec^2 x$$

Let $$f(x) = \sin^2 x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{\sin^2(x + h) - \sin^2 x}{h}$$

Use the identity $$\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$$ with $$A = x + h$$ and $$B = x$$:

$$\sin^2(x + h) - \sin^2 x = \sin(2x + h)\sin h$$

So

$$f'(x) = \lim_{h \to 0}\frac{\sin(2x + h)\sin h}{h} = \lim_{h \to 0}\sin(2x + h) \cdot \frac{\sin h}{h}$$

As $$h \to 0$$: $$\sin(2x + h) \to \sin 2x$$ and $$\dfrac{\sin h}{h} \to 1$$. Hence

$$f'(x) = \sin 2x \cdot 1 = \sin 2x$$

Let $$f(x) = x^2 - 2$$. By the first principle, the derivative at $$x = 10$$ is

$$f'(10) = \lim_{h \to 0}\frac{f(10 + h) - f(10)}{h}$$

Compute the two values:

$$f(10 + h) = (10 + h)^2 - 2 = 100 + 20h + h^2 - 2 = 98 + 20h + h^2$$

$$f(10) = (10)^2 - 2 = 98$$

So

$$f(10 + h) - f(10) = 20h + h^2 = h(20 + h)$$

Therefore

$$f'(10) = \lim_{h \to 0}\frac{h(20 + h)}{h} = \lim_{h \to 0}(20 + h) = 20$$

Let $$f(x) = x$$. By the first principle, the derivative at $$x = 1$$ is

$$f'(1) = \lim_{h \to 0}\frac{f(1 + h) - f(1)}{h}$$

Here $$f(1 + h) = 1 + h$$ and $$f(1) = 1$$, so

$$f'(1) = \lim_{h \to 0}\frac{(1 + h) - 1}{h} = \lim_{h \to 0}\frac{h}{h} = \lim_{h \to 0} 1 = 1$$

Let $$f(x) = 99x$$. By the first principle, the derivative at $$x = 100$$ is

$$f'(100) = \lim_{h \to 0}\frac{f(100 + h) - f(100)}{h}$$

Here $$f(100 + h) = 99(100 + h) = 9900 + 99h$$ and $$f(100) = 99(100) = 9900$$, so

$$f(100 + h) - f(100) = 99h$$

Therefore

$$f'(100) = \lim_{h \to 0}\frac{99h}{h} = \lim_{h \to 0} 99 = 99$$

Differentiate term by term using the power rule $$\dfrac{d}{dx}x^n = nx^{n-1}$$. A typical term is $$\dfrac{x^n}{n}$$, whose derivative is

$$\frac{d}{dx}\left(\frac{x^n}{n}\right) = \frac{n x^{n-1}}{n} = x^{n-1}$$

Applying this to every term (and noting the constant $$1$$ has derivative $$0$$):

$$f'(x) = x^{99} + x^{98} + \ldots + x^2 + x + 1$$

This is a sum of $$100$$ terms (the powers $$x^{99}$$ down to $$x^0 = 1$$).

Value at $$x = 1$$: every power of $$1$$ equals $$1$$, so

$$f'(1) = \underbrace{1 + 1 + \ldots + 1}_{100 \text{ terms}} = 100$$

Value at $$x = 0$$: every term with a positive power of $$x$$ vanishes, leaving only the last term $$1$$:

$$f'(0) = 0 + 0 + \ldots + 0 + 1 = 1$$

Hence

$$f'(1) = 100 = 100 \times 1 = 100\,f'(0)$$

which proves $$f'(1) = 100\,f'(0)$$.

Let $$f(x) = x^n + ax^{n-1} + a^2 x^{n-2} + \ldots + a^{n-1}x + a^n$$. Here $$a$$ is a constant, so each power $$a^k$$ is just a constant coefficient.

Differentiate term by term using $$\dfrac{d}{dx}x^m = m x^{m-1}$$:

$$\frac{d}{dx}\big(x^n\big) = n x^{n-1}$$

$$\frac{d}{dx}\big(a x^{n-1}\big) = a(n - 1)x^{n-2}$$

$$\frac{d}{dx}\big(a^2 x^{n-2}\big) = a^2(n - 2)x^{n-3}$$

and so on, while the last term $$a^n$$ is a constant with derivative $$0$$. Collecting all the terms:

$$f'(x) = n x^{n-1} + a(n - 1)x^{n-2} + a^2(n - 2)x^{n-3} + \ldots + a^{n-1}$$

Let $$f(x) = \dfrac{x^n - a^n}{x - a}$$. Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = x^n - a^n, \quad u' = nx^{n-1}, \qquad v = x - a, \quad v' = 1$$

Then

$$f'(x) = \frac{nx^{n-1}(x - a) - (x^n - a^n)(1)}{(x - a)^2}$$

Expand the numerator:

$$nx^{n-1}(x - a) = nx^n - nax^{n-1}$$

$$\text{Numerator} = nx^n - nax^{n-1} - x^n + a^n = (n - 1)x^n - nax^{n-1} + a^n$$

Therefore

$$f'(x) = \frac{(n - 1)x^n - nax^{n-1} + a^n}{(x - a)^2}$$

Let $$f(x) = \cos x$$. By the first principle,

$$f'(x) = \lim_{h \to 0}\frac{\cos(x + h) - \cos x}{h}$$

Use the difference-to-product identity $$\cos C - \cos D = -2\sin\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$$ with $$C = x + h$$ and $$D = x$$:

$$\cos(x + h) - \cos x = -2\sin\left(x + \frac{h}{2}\right)\sin\frac{h}{2}$$

So

$$f'(x) = \lim_{h \to 0}\frac{-2\sin\left(x + \frac{h}{2}\right)\sin\frac{h}{2}}{h} = \lim_{h \to 0}\left[-\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\frac{h}{2}}{\frac{h}{2}}\right]$$

As $$h \to 0$$: $$\sin\left(x + \dfrac{h}{2}\right) \to \sin x$$ and $$\dfrac{\sin(h/2)}{h/2} \to 1$$. Therefore

$$f'(x) = -\sin x \cdot 1 = -\sin x$$

Let $$f(x) = x + a$$, where $$a$$ is a constant. Differentiate term by term: the derivative of $$x$$ is $$1$$, and the derivative of the constant $$a$$ is $$0$$.

$$\frac{d}{dx}(x + a) = 1 + 0 = 1$$

It is simplest to expand the product first:

$$(px + q)\left(\frac{r}{x} + s\right) = px\cdot\frac{r}{x} + px\cdot s + q\cdot\frac{r}{x} + q\cdot s = pr + psx + \frac{qr}{x} + qs$$

So the function is

$$psx + \frac{qr}{x} + (pr + qs)$$

where $$pr + qs$$ is a constant. Differentiate term by term, using $$\dfrac{d}{dx}\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^2}$$:

$$\frac{d}{dx}(psx) = ps, \qquad \frac{d}{dx}\left(\frac{qr}{x}\right) = -\frac{qr}{x^2}, \qquad \frac{d}{dx}(pr + qs) = 0$$

Therefore

$$\frac{d}{dx}\left[(px + q)\left(\frac{r}{x} + s\right)\right] = ps - \frac{qr}{x^2}$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = ax + b$$ and $$v = (cx + d)^2$$.

Here $$u' = a$$. For $$v$$, expand $$(cx + d)^2 = c^2 x^2 + 2cdx + d^2$$, whose derivative is

$$v' = 2c^2 x + 2cd = 2c(cx + d)$$

Then

$$\frac{d}{dx}\big[(ax + b)(cx + d)^2\big] = a(cx + d)^2 + (ax + b)\cdot 2c(cx + d)$$

Take the common factor $$(cx + d)$$:

$$= (cx + d)\big[a(cx + d) + 2c(ax + b)\big]$$

Simplify the bracket: $$a(cx + d) + 2c(ax + b) = acx + ad + 2acx + 2bc = 3acx + ad + 2bc$$.

$$\frac{d}{dx}\big[(ax + b)(cx + d)^2\big] = (cx + d)(3acx + ad + 2bc)$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = ax + b, \quad u' = a, \qquad v = cx + d, \quad v' = c$$

Then

$$\frac{d}{dx}\left(\frac{ax + b}{cx + d}\right) = \frac{a(cx + d) - (ax + b)c}{(cx + d)^2}$$

Expand the numerator:

$$a(cx + d) - c(ax + b) = acx + ad - acx - bc = ad - bc$$

Therefore

$$\frac{d}{dx}\left(\frac{ax + b}{cx + d}\right) = \frac{ad - bc}{(cx + d)^2}$$

First simplify the compound fraction by multiplying numerator and denominator by $$x$$:

$$\frac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}} = \frac{x\left(1 + \dfrac{1}{x}\right)}{x\left(1 - \dfrac{1}{x}\right)} = \frac{x + 1}{x - 1}$$

Now apply the quotient rule with $$u = x + 1,\, u' = 1$$ and $$v = x - 1,\, v' = 1$$:

$$\frac{d}{dx}\left(\frac{x + 1}{x - 1}\right) = \frac{(1)(x - 1) - (x + 1)(1)}{(x - 1)^2}$$

Simplify the numerator: $$(x - 1) - (x + 1) = -2$$.

$$\frac{d}{dx}\left(\frac{x + 1}{x - 1}\right) = \frac{-2}{(x - 1)^2}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = 1, \quad u' = 0, \qquad v = ax^2 + bx + c, \quad v' = 2ax + b$$

Then

$$\frac{d}{dx}\left(\frac{1}{ax^2 + bx + c}\right) = \frac{0\cdot(ax^2 + bx + c) - 1\cdot(2ax + b)}{(ax^2 + bx + c)^2}$$

$$= \frac{-(2ax + b)}{(ax^2 + bx + c)^2}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = ax + b, \quad u' = a, \qquad v = px^2 + qx + r, \quad v' = 2px + q$$

Then

$$\frac{d}{dx}\left(\frac{ax + b}{px^2 + qx + r}\right) = \frac{a(px^2 + qx + r) - (ax + b)(2px + q)}{(px^2 + qx + r)^2}$$

Expand the numerator:

$$a(px^2 + qx + r) = apx^2 + aqx + ar$$

$$(ax + b)(2px + q) = 2apx^2 + aqx + 2bpx + bq$$

$$\text{Numerator} = (apx^2 + aqx + ar) - (2apx^2 + aqx + 2bpx + bq) = -apx^2 - 2bpx + ar - bq$$

Therefore

$$\frac{d}{dx}\left(\frac{ax + b}{px^2 + qx + r}\right) = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = px^2 + qx + r, \quad u' = 2px + q, \qquad v = ax + b, \quad v' = a$$

Then

$$\frac{d}{dx}\left(\frac{px^2 + qx + r}{ax + b}\right) = \frac{(2px + q)(ax + b) - (px^2 + qx + r)a}{(ax + b)^2}$$

Expand the numerator:

$$(2px + q)(ax + b) = 2apx^2 + 2bpx + aqx + bq$$

$$a(px^2 + qx + r) = apx^2 + aqx + ar$$

$$\text{Numerator} = (2apx^2 + 2bpx + aqx + bq) - (apx^2 + aqx + ar) = apx^2 + 2bpx + bq - ar$$

Therefore

$$\frac{d}{dx}\left(\frac{px^2 + qx + r}{ax + b}\right) = \frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2}$$

Rewrite the powers of $$x$$ with negative exponents:

$$\frac{a}{x^4} - \frac{b}{x^2} + \cos x = a x^{-4} - b x^{-2} + \cos x$$

Differentiate term by term, using the power rule and $$\dfrac{d}{dx}\cos x = -\sin x$$:

$$\frac{d}{dx}\big(a x^{-4}\big) = a(-4)x^{-5} = -4a x^{-5}$$

$$\frac{d}{dx}\big(b x^{-2}\big) = b(-2)x^{-3} = -2b x^{-3}$$

$$\frac{d}{dx}(\cos x) = -\sin x$$

Therefore

$$\frac{d}{dx}\left(\frac{a}{x^4} - \frac{b}{x^2} + \cos x\right) = -4a x^{-5} - (-2b x^{-3}) - \sin x = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x$$

Write the square root as a power: $$\sqrt{x} = x^{1/2}$$, so the function is $$4x^{1/2} - 2$$.

Differentiate term by term using the power rule $$\dfrac{d}{dx}x^n = nx^{n-1}$$, valid here with $$n = \dfrac{1}{2}$$. The constant $$2$$ has derivative $$0$$.

$$\frac{d}{dx}\big(4x^{1/2}\big) = 4 \cdot \frac{1}{2} x^{1/2 - 1} = 2x^{-1/2}$$

Therefore

$$\frac{d}{dx}\big(4\sqrt{x} - 2\big) = 2x^{-1/2} = \frac{2}{\sqrt{x}}$$

Let $$f(x) = (ax + b)^n$$. We work from the first principle.

$$f'(x) = \lim_{h \to 0}\frac{(a(x + h) + b)^n - (ax + b)^n}{h}$$

Write $$u = ax + b$$, so $$a(x + h) + b = ax + b + ah = u + ah$$. Expand $$(u + ah)^n$$ by the binomial theorem:

$$(u + ah)^n = u^n + n\,u^{n-1}(ah) + \binom{n}{2}u^{n-2}(ah)^2 + \ldots$$

Subtracting $$u^n$$ and dividing by $$h$$:

$$\frac{(u + ah)^n - u^n}{h} = n\,u^{n-1}a + \binom{n}{2}u^{n-2}a^2 h + \ldots$$

Every term after the first still contains a factor $$h$$, so it vanishes as $$h \to 0$$. Hence

$$f'(x) = n\,u^{n-1}a = na(ax + b)^{n-1}$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = (ax + b)^n$$ and $$v = (cx + d)^m$$.

From Question 12, the derivatives of these powers are

$$u' = na(ax + b)^{n-1}, \qquad v' = mc(cx + d)^{m-1}$$

Then

$$\frac{d}{dx}\big[(ax + b)^n (cx + d)^m\big] = na(ax + b)^{n-1}(cx + d)^m + (ax + b)^n \cdot mc(cx + d)^{m-1}$$

Take out the common factor $$(ax + b)^{n-1}(cx + d)^{m-1}$$:

$$= (ax + b)^{n-1}(cx + d)^{m-1}\big[na(cx + d) + mc(ax + b)\big]$$

Expand $$\sin(x + a)$$ using the addition formula. Since $$a$$ is a constant, $$\cos a$$ and $$\sin a$$ are constants:

$$\sin(x + a) = \sin x \cos a + \cos x \sin a$$

Differentiate term by term, using $$\dfrac{d}{dx}\sin x = \cos x$$ and $$\dfrac{d}{dx}\cos x = -\sin x$$:

$$\frac{d}{dx}\sin(x + a) = \cos a \cdot \cos x + \sin a \cdot (-\sin x) = \cos x \cos a - \sin x \sin a$$

The right side is the expansion of $$\cos(x + a)$$, so

$$\frac{d}{dx}\sin(x + a) = \cos(x + a)$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = \csc x$$ and $$v = \cot x$$. Recall

$$\frac{d}{dx}\csc x = -\csc x \cot x, \qquad \frac{d}{dx}\cot x = -\csc^2 x$$

Then

$$\frac{d}{dx}(\csc x \cot x) = (-\csc x \cot x)(\cot x) + (\csc x)(-\csc^2 x)$$

$$= -\csc x \cot^2 x - \csc^3 x$$

Take the common factor $$-\csc x$$:

$$= -\csc x\,(\cot^2 x + \csc^2 x)$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = \cos x, \quad u' = -\sin x, \qquad v = 1 + \sin x, \quad v' = \cos x$$

Then

$$\frac{d}{dx}\left(\frac{\cos x}{1 + \sin x}\right) = \frac{(-\sin x)(1 + \sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}$$

Expand the numerator:

$$-\sin x - \sin^2 x - \cos^2 x = -\sin x - (\sin^2 x + \cos^2 x) = -\sin x - 1$$

So

$$\frac{d}{dx}\left(\frac{\cos x}{1 + \sin x}\right) = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = \frac{-1}{1 + \sin x}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = \sin x + \cos x, \quad u' = \cos x - \sin x$$

$$v = \sin x - \cos x, \quad v' = \cos x + \sin x$$

Then

$$\frac{d}{dx}\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}$$

Evaluate each product. Note $$(\cos x - \sin x)(\sin x - \cos x) = -(\sin x - \cos x)^2 = -(1 - 2\sin x\cos x)$$, and $$(\sin x + \cos x)^2 = 1 + 2\sin x\cos x$$.

$$\text{Numerator} = -(1 - 2\sin x\cos x) - (1 + 2\sin x\cos x) = -1 + 2\sin x\cos x - 1 - 2\sin x\cos x = -2$$

Therefore

$$\frac{d}{dx}\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right) = \frac{-2}{(\sin x - \cos x)^2}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = \sec x - 1, \quad u' = \sec x \tan x, \qquad v = \sec x + 1, \quad v' = \sec x \tan x$$

Then

$$\frac{d}{dx}\left(\frac{\sec x - 1}{\sec x + 1}\right) = \frac{\sec x \tan x\,(\sec x + 1) - (\sec x - 1)\,\sec x \tan x}{(\sec x + 1)^2}$$

Take the common factor $$\sec x \tan x$$ in the numerator:

$$= \frac{\sec x \tan x\,\big[(\sec x + 1) - (\sec x - 1)\big]}{(\sec x + 1)^2} = \frac{\sec x \tan x \cdot 2}{(\sec x + 1)^2}$$

Therefore

$$\frac{d}{dx}\left(\frac{\sec x - 1}{\sec x + 1}\right) = \frac{2\sec x \tan x}{(\sec x + 1)^2}$$

Let $$f(x) = \sin^n x = (\sin x)^n$$. We prove, by induction on $$n$$, that

$$\frac{d}{dx}\sin^n x = n\sin^{n-1}x\cos x$$

Base case ($$n = 1$$). $$\dfrac{d}{dx}\sin x = \cos x$$, which agrees with $$1 \cdot \sin^0 x \cos x = \cos x$$.

Inductive step. Assume the result holds for some $$n = k$$, that is $$\dfrac{d}{dx}\sin^k x = k\sin^{k-1}x\cos x$$. Write $$\sin^{k+1}x = \sin^k x \cdot \sin x$$ and apply the product rule:

$$\frac{d}{dx}\sin^{k+1}x = \left(\frac{d}{dx}\sin^k x\right)\sin x + \sin^k x\left(\frac{d}{dx}\sin x\right)$$

$$= \big(k\sin^{k-1}x\cos x\big)\sin x + \sin^k x\cos x = k\sin^k x\cos x + \sin^k x\cos x$$

$$= (k + 1)\sin^k x\cos x$$

This is the formula for $$n = k + 1$$. By induction it holds for every positive integer $$n$$:

$$\frac{d}{dx}\sin^n x = n\sin^{n-1}x\cos x$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = a + b\sin x, \quad u' = b\cos x, \qquad v = c + d\cos x, \quad v' = -d\sin x$$

Then

$$\frac{d}{dx}\left(\frac{a + b\sin x}{c + d\cos x}\right) = \frac{b\cos x\,(c + d\cos x) - (a + b\sin x)(-d\sin x)}{(c + d\cos x)^2}$$

$$= \frac{b\cos x\,(c + d\cos x) + d\sin x\,(a + b\sin x)}{(c + d\cos x)^2}$$

Expand the numerator:

$$bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x$$

$$= bc\cos x + ad\sin x + bd(\cos^2 x + \sin^2 x) = bc\cos x + ad\sin x + bd$$

Therefore

$$\frac{d}{dx}\left(\frac{a + b\sin x}{c + d\cos x}\right) = \frac{bc\cos x + ad\sin x + bd}{(c + d\cos x)^2}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = \sin(x + a), \quad u' = \cos(x + a), \qquad v = \cos x, \quad v' = -\sin x$$

(The derivative $$\dfrac{d}{dx}\sin(x + a) = \cos(x + a)$$ was established in Question 14.)

$$\frac{d}{dx}\left(\frac{\sin(x + a)}{\cos x}\right) = \frac{\cos(x + a)\cos x - \sin(x + a)(-\sin x)}{\cos^2 x}$$

$$= \frac{\cos(x + a)\cos x + \sin(x + a)\sin x}{\cos^2 x}$$

The numerator is $$\cos\big((x + a) - x\big) = \cos a$$. Therefore

$$\frac{d}{dx}\left(\frac{\sin(x + a)}{\cos x}\right) = \frac{\cos a}{\cos^2 x}$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = x^4$$ and $$v = 5\sin x - 3\cos x$$.

$$u' = 4x^3, \qquad v' = 5\cos x - 3(-\sin x) = 5\cos x + 3\sin x$$

Then

$$\frac{d}{dx}\big[x^4(5\sin x - 3\cos x)\big] = 4x^3(5\sin x - 3\cos x) + x^4(5\cos x + 3\sin x)$$

Take out the common factor $$x^3$$:

$$= x^3\big[4(5\sin x - 3\cos x) + x(5\cos x + 3\sin x)\big]$$

$$= x^3\big(20\sin x - 12\cos x + 5x\cos x + 3x\sin x\big)$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = x^2 + 1$$ and $$v = \cos x$$.

$$u' = 2x, \qquad v' = -\sin x$$

Then

$$\frac{d}{dx}\big[(x^2 + 1)\cos x\big] = 2x\cos x + (x^2 + 1)(-\sin x)$$

$$= 2x\cos x - (x^2 + 1)\sin x$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = ax^2 + \sin x$$ and $$v = p + q\cos x$$.

$$u' = 2ax + \cos x, \qquad v' = -q\sin x$$

Then

$$\frac{d}{dx}\big[(ax^2 + \sin x)(p + q\cos x)\big] = (2ax + \cos x)(p + q\cos x) + (ax^2 + \sin x)(-q\sin x)$$

$$= (2ax + \cos x)(p + q\cos x) - q\sin x\,(ax^2 + \sin x)$$

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = x + \cos x$$ and $$v = x - \tan x$$.

$$u' = 1 - \sin x, \qquad v' = 1 - \sec^2 x$$

Then

$$\frac{d}{dx}\big[(x + \cos x)(x - \tan x)\big] = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = 4x + 5\sin x, \quad u' = 4 + 5\cos x$$

$$v = 3x + 7\cos x, \quad v' = 3 - 7\sin x$$

Then

$$\frac{d}{dx}\left(\frac{4x + 5\sin x}{3x + 7\cos x}\right) = \frac{(4 + 5\cos x)(3x + 7\cos x) - (4x + 5\sin x)(3 - 7\sin x)}{(3x + 7\cos x)^2}$$

Expand each product in the numerator:

$$(4 + 5\cos x)(3x + 7\cos x) = 12x + 28\cos x + 15x\cos x + 35\cos^2 x$$

$$(4x + 5\sin x)(3 - 7\sin x) = 12x - 28x\sin x + 15\sin x - 35\sin^2 x$$

Subtract the second from the first. The $$12x$$ terms cancel, and $$35\cos^2 x + 35\sin^2 x = 35$$:

$$\text{Numerator} = 28\cos x + 15x\cos x + 28x\sin x - 15\sin x + 35$$

Therefore

$$\frac{d}{dx}\left(\frac{4x + 5\sin x}{3x + 7\cos x}\right) = \frac{35 + 28\cos x + 15x\cos x + 28x\sin x - 15\sin x}{(3x + 7\cos x)^2}$$

Here $$\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$$ is a constant, so it can be taken outside the derivative:

$$\frac{d}{dx}\left(\frac{x^2 \cos\frac{\pi}{4}}{\sin x}\right) = \cos\frac{\pi}{4} \cdot \frac{d}{dx}\left(\frac{x^2}{\sin x}\right)$$

Apply the quotient rule to $$\dfrac{x^2}{\sin x}$$ with $$u = x^2,\, u' = 2x$$ and $$v = \sin x,\, v' = \cos x$$:

$$\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{2x\sin x - x^2\cos x}{\sin^2 x}$$

Therefore

$$\frac{d}{dx}\left(\frac{x^2 \cos\frac{\pi}{4}}{\sin x}\right) = \cos\frac{\pi}{4} \cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} = \frac{2x\sin x - x^2\cos x}{\sqrt{2}\,\sin^2 x}$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with

$$u = x, \quad u' = 1, \qquad v = 1 + \tan x, \quad v' = \sec^2 x$$

Then

$$\frac{d}{dx}\left(\frac{x}{1 + \tan x}\right) = \frac{1\cdot(1 + \tan x) - x\sec^2 x}{(1 + \tan x)^2}$$

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

Apply the product rule $$(uv)' = u'v + uv'$$ with $$u = x + \sec x$$ and $$v = x - \tan x$$.

$$u' = 1 + \sec x \tan x, \qquad v' = 1 - \sec^2 x$$

Then

$$\frac{d}{dx}\big[(x + \sec x)(x - \tan x)\big] = (1 + \sec x \tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)$$

Apply the quotient rule $$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$$ with $$u = x$$ and $$v = \sin^n x$$.

$$u' = 1, \qquad v' = n\sin^{n-1}x\cos x$$

(The derivative $$\dfrac{d}{dx}\sin^n x = n\sin^{n-1}x\cos x$$ was established in Question 19.)

Then

$$\frac{d}{dx}\left(\frac{x}{\sin^n x}\right) = \frac{1\cdot\sin^n x - x\cdot n\sin^{n-1}x\cos x}{(\sin^n x)^2} = \frac{\sin^n x - nx\sin^{n-1}x\cos x}{\sin^{2n} x}$$

Take the common factor $$\sin^{n-1}x$$ out of the numerator:

$$= \frac{\sin^{n-1}x\,(\sin x - nx\cos x)}{\sin^{2n} x} = \frac{\sin x - nx\cos x}{\sin^{n+1} x}$$