Let $$f(x) = x|x|$$, $$g(x) = \sin x$$ and $$h(x) = (g \circ f)(x)$$. Then:

We are given three real-valued functions.

First, $$f(x)=x\,|x|$$. By analysing the sign of $$x$$, we can write this function in an explicit piece-wise form:

For $$x\ge 0$$ we have $$|x|=x$$, so $$ f(x)=x\cdot x=x^{2}\qquad (x\ge 0). $$

For $$x<0$$ we have $$|x|=-x$$, so $$ f(x)=x\cdot(-x)=-x^{2}\qquad (x<0). $$

Hence $$ f(x)= \begin{cases} x^{2}, & x\ge 0,\\[4pt] -\,x^{2}, & x<0. \end{cases} $$

Second, $$g(x)=\sin x$$.

We define $$h(x)=(g\circ f)(x)=g\bigl(f(x)\bigr)=\sin\bigl(f(x)\bigr).$$ Substituting the above piece-wise form of $$f(x)$$, we obtain

$$ h(x)= \begin{cases} \sin\bigl(x^{2}\bigr), & x\ge 0,\\[4pt] \sin\bigl(-x^{2}\bigr), & x<0. \end{cases} $$

Because $$\sin(-\theta)=-\,\sin\theta$$, this can be rewritten as

$$ h(x)= \begin{cases} \sin(x^{2}), & x\ge 0,\\[4pt] -\,\sin(x^{2}), & x<0. \end{cases} $$

Now we examine differentiability at $$x=0$$ step by step.

Step 1: Find $$f'(x)$$.

For $$x\ne 0$$ the derivative of $$f(x)=x|x|$$ is found directly:

• For $$x>0$$, $$f(x)=x^{2}$$, so $$f'(x)=2x.$$
• For $$x<0$$, $$f(x)=-x^{2}$$, so $$f'(x)=-\,2x.$$

At $$x=0$$ we evaluate the limit definition of derivative:

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

Collecting these results,

$$ f'(x)= \begin{cases} 2x, & x>0,\\[4pt] 0, & x=0,\\[4pt] -\,2x, & x<0. \end{cases} $$

Step 2: Find $$h'(x)$$ using the Chain Rule.

The Chain Rule states: if $$y=g(f(x))$$ then $$\displaystyle\frac{dy}{dx}=g'\bigl(f(x)\bigr)\,f'(x).$$ Applying it to $$h(x)=\sin\bigl(f(x)\bigr)$$ gives

$$ h'(x)=\cos\bigl(f(x)\bigr)\;f'(x). $$

Because $$f(x)$$ is even and $$\cos(\theta)$$ is also even, it is convenient to keep $$\cos(f(x))$$ unchanged. Substituting the earlier piece-wise forms of $$f(x)$$ and $$f'(x)$$, we obtain for $$x\ne 0$$:

• For $$x>0$$, $$ h'(x)=\cos\!\bigl(x^{2}\bigr)\;\cdot\;2x =2x\,\cos(x^{2}). $$
• For $$x<0$$, $$ h'(x)=\cos\!\bigl(-x^{2}\bigr)\;\cdot\;(-2x) =\cos(x^{2})\,(-2x) =-\,2x\,\cos(x^{2}), $$ because $$\cos(-\theta)=\cos\theta.$$

At $$x=0$$ we use $$f'(0)=0$$ together with $$\cos\bigl(f(0)\bigr)=\cos 0=1$$, giving

$$ h'(0)=\cos(0)\cdot f'(0)=1\cdot 0=0. $$

Thus

$$ h'(x)= \begin{cases} 2x\,\cos(x^{2}), & x>0,\\[4pt] 0, & x=0,\\[4pt] -\,2x\,\cos(x^{2}), & x<0. \end{cases} $$

Step 3: Continuity of $$h'(x)$$ at $$x=0$$.

We compute the right-hand and left-hand limits of $$h'(x)$$ as $$x\to 0$$.

For $$x\to 0^{+}$$, $$ \lim_{x\to 0^{+}}h'(x)=\lim_{x\to 0^{+}}2x\,\cos(x^{2})=2\cdot 0\cdot 1=0. $$

For $$x\to 0^{-}$$, $$ \lim_{x\to 0^{-}}h'(x)=\lim_{x\to 0^{-}}(-2x)\,\cos(x^{2})=(-2)\cdot 0\cdot 1=0. $$

Both one-sided limits equal the value $$h'(0)=0,$$ so $$h'(x)$$ is continuous at $$x=0$$.

Step 4: Differentiate $$h'(x)$$ once more to test the differentiability of $$h'(x)$$ at $$x=0$$.

For $$x>0$$ we differentiate $$h'(x)=2x\,\cos(x^{2})$$ using the Product Rule $$\dfrac{d}{dx}[u\,v]=u'\,v+u\,v'$$ and the Chain Rule $$\dfrac{d}{dx}\cos(x^{2})=-\sin(x^{2})\cdot 2x$$:

$$ \begin{aligned} h''(x) &=\frac{d}{dx}\!\bigl(2x\,\cos(x^{2})\bigr)\\[4pt] &=2\cos(x^{2})+2x\bigl(-\sin(x^{2})\cdot 2x\bigr)\\[4pt] &=2\cos(x^{2})-4x^{2}\sin(x^{2}),\qquad x>0. \end{aligned} $$

For $$x<0$$, $$h'(x)=-2x\,\cos(x^{2}).$$ Differentiating this expression gives

$$ \begin{aligned} h''(x) &=\frac{d}{dx}\!\bigl(-2x\,\cos(x^{2})\bigr)\\[4pt] &=-2\cos(x^{2})+(-2x)\bigl(-\sin(x^{2})\cdot 2x\bigr)\\[4pt] &=-2\cos(x^{2})+4x^{2}\sin(x^{2}),\qquad x<0. \end{aligned} $$

Now we find the one-sided limits of $$h''(x)$$ as $$x\to 0$$.

For $$x\to 0^{+}$$, the term $$-4x^{2}\sin(x^{2})\to 0$$ and $$\cos(x^{2})\to 1,$$ so $$ \lim_{x\to 0^{+}}h''(x)=2\cdot 1=2. $$

For $$x\to 0^{-}$$, the term $$4x^{2}\sin(x^{2})\to 0$$ and again $$\cos(x^{2})\to 1,$$ so $$ \lim_{x\to 0^{-}}h''(x)=-2\cdot 1=-2. $$

The right-hand limit equals $$2$$ while the left-hand limit equals $$-2$$. Because the two limits are unequal, $$h''(0)$$ does not exist. Therefore $$h'(x)$$ is not differentiable at $$x=0$$.

We have shown that

• $$h'(x)$$ exists for every $$x$$,
• $$h'(x)$$ is continuous at $$x=0$$,
• but $$h'(x)$$ is not differentiable at $$x=0$$.

This matches the description in Option C: “$$h'(x)$$ is continuous at $$x = 0$$ but it is not differentiable at $$x = 0$$.”

Hence, the correct answer is Option C.