Differentiability of Piecewise and Composite Functions

To determine the correct statements, we analyze each option individually by checking differentiability at $x = 0$.

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Option A: Differentiability of $f(x)$ at $x = 0$

The function $f(x)$ is defined as:

$$f(x) = \begin{cases} \frac{x}{|x|} g(x), & x \neq 0 \\ 0, & x = 0 \end{cases}$$

By the definition of the derivative at $x = 0$: $f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{\frac{x}{|x|} g(x) - 0}{x} = \lim_{x \to 0} \frac{g(x)}{|x|}$

We evaluate the right-hand derivative ($\text{RHD}$) and left-hand derivative ($\text{LHD}$):

• Right-hand derivative ($\text{RHD}$): $\text{RHD} = \lim_{x \to 0^+} \frac{g(x)}{x} = g'(0) = 0$

• Left-hand derivative ($\text{LHD}$): $\text{LHD} = \lim_{x \to 0^-} \frac{g(x)}{-x} = - \lim_{x \to 0^-} \frac{g(x)}{x} = -g'(0) = 0$

Since $\text{RHD} = \text{LHD} = 0$, $f(x)$ is differentiable at $x = 0$, and $f'(0) = 0$.

Thus, Option A is correct.

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Option B: Differentiability of $h(x)$ at $x = 0$

The function $h(x) = e^{|x|}$. Let us compute the one-sided derivatives at $x = 0$:

• Right-hand derivative ($\text{RHD}$): $\text{RHD} = \lim_{x \to 0^+} \frac{e^{|x|} - e^0}{x} = \lim_{x \to 0^+} \frac{e^x - 1}{x} = 1$

• Left-hand derivative ($\text{LHD}$): $\text{LHD} = \lim_{x \to 0^-} \frac{e^{|x|} - e^0}{x} = \lim_{x \to 0^-} \frac{e^{-x} - 1}{x} = -1$

Since $\text{RHD} \neq \text{LHD}$, $h(x)$ is not differentiable at $x = 0$.

Thus, Option B is incorrect.

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Option C: Differentiability of $(f \circ h)(x)$ at $x = 0$

We have $(f \circ h)(x) = f(h(x)) = f(e^{|x|})$.

Since $e^{|x|} > 0$ for all $x \in \mathbb{R}$, we have $|e^{|x|}| = e^{|x|}$. Therefore, for $x \in \mathbb{R}$: $f(e^{|x|}) = \frac{e^{|x|}}{|e^{|x|}|} g(e^{|x|}) = g(e^{|x|})$

Now, let $P(x) = (f \circ h)(x) = g(e^{|x|})$. Note that $P(0) = g(e^0) = g(1)$.

Let's test the one-sided derivatives of $P(x)$ at $x = 0$:

• Right-hand derivative ($\text{RHD}$): $\text{RHD} = \lim_{x \to 0^+} \frac{g(e^x) - g(1)}{x}$ Using L'Hôpital's Rule: $\text{RHD} = \lim_{x \to 0^+} \frac{g'(e^x) \cdot e^x}{1} = g'(1) \cdot 1 = g'(1)$

• Left-hand derivative ($\text{LHD}$): $\text{LHD} = \lim_{x \to 0^-} \frac{g(e^{-x}) - g(1)}{x}$ Using L'Hôpital's Rule: $\text{LHD} = \lim_{x \to 0^-} \frac{g'(e^{-x}) \cdot (-e^{-x})}{1} = -g'(1)$

Given that $g'(1) \neq 0$, it follows that $g'(1) \neq -g'(1)$, so $\text{RHD} \neq \text{LHD}$. Hence, $(f \circ h)(x)$ is not differentiable at $x = 0$.

Thus, Option C is incorrect.

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Option D: Differentiability of $(h \circ f)(x)$ at $x = 0$

We have $(h \circ f)(x) = h(f(x)) = e^{|f(x)|}$. Note that $(h \circ f)(0) = e^{|f(0)|} = e^0 = 1$.

Let $Q(x) = (h \circ f)(x) = e^{|f(x)|}$. The derivative at $x = 0$ is given by: $Q'(0) = \lim_{x \to 0} \frac{e^{|f(x)|} - 1}{x}$

For $x \neq 0$, we have: $|f(x)| = \left| \frac{x}{|x|} g(x) \right| = |g(x)|$

Thus, $Q'(0) = \lim_{x \to 0} \frac{e^{|g(x)|} - 1}{x}$

For $g(x) \neq 0$, we can write: $\lim_{x \to 0} \frac{e^{|g(x)|} - 1}{x} = \lim_{x \to 0} \left( \frac{e^{|g(x)|} - 1}{|g(x)|} \cdot \frac{|g(x)|}{x} \right)$

Since $g(0) = 0$ and $g$ is continuous, $|g(x)| \to 0$ as $x \to 0$. Hence: $\lim_{x \to 0} \frac{e^{|g(x)|} - 1}{|g(x)|} = 1$

Next, we evaluate $\lim_{x \to 0} \frac{|g(x)|}{x}$: $\lim_{x \to 0} \left| \frac{g(x)}{x} \right| = |g'(0)| = 0 \implies \lim_{x \to 0} \frac{|g(x)|}{x} = 0$

Therefore: $Q'(0) = 1 \cdot 0 = 0$

Since the limit exists and equals $0$, $(h \circ f)(x)$ is differentiable at $x = 0$.

Thus, Option D is correct.

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Conclusion

The correct options are A and D.