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(D) Given $g(x) = h(e^x) \cdot e^{h(x)}$.Applying the product rule for differentiation,we get:$g^{\prime}(x) = h(e^x) \cdot \frac{d}{dx}(e^{h(x)}) + e

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$4$

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(D) Given $g(x) = h(e^x) \cdot e^{h(x)}$.Applying the product rule for differentiation,we get:$g^{\prime}(x) = h(e^x) \cdot \frac{d}{dx}(e^{h(x)}) + e^{h(x)} \cdot \frac{d}{dx}(h(e^x))$$g^{\prime}(x) = h(e^x) \cdot e^{h(x)} \cdot h^{\prime}(x) + e^{h(x)} \cdot h^{\prime}(e^x) \cdot e^x$Now,substitute $x = 0$ into the expression:$g^{\prime}(0) = h(e^0) \cdot e^{h(0)} \cdot h^{\prime}(0) + e^{h(0)} \cdot h^{\prime}(e^0) \cdot e^0$$g^{\prime}(0) = h(1) \cdot e^{h(0)} \cdot h^{\prime}(0) + e^{h(0)} \cdot h^{\prime}(1) \cdot 1$Given $h(0)=0$,$h(1)=1$,and $h^{\prime}(0)=h^{\prime}(1)=2$:$g^{\prime}(0) = (1) \cdot e^0 \cdot (2) + e^0 \cdot (2) \cdot 1$$g^{\prime}(0) = 1 \cdot 1 \cdot 2 + 1 \cdot 2 \cdot 1$$g^{\prime}(0) = 2 + 2 = 4$.

Suppose for a differentiable function $h$,$h(0)=0$,$h(1)=1$ and $h^{\prime}(0)=h^{\prime}(1)=2$. If $g(x)=h(e^{x}) e^{h(x)}$,then $g^{\prime}(0)$ is equal to:

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