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

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Aicharya · Accepted Answer

$\begin{aligned} & g(x)=h\left(e^xight) \cdot e^{h(x)} \ & g^{\prime}(x)=h\left(e^xight) \cdot e^{h(x)} \cdot h^{\prime}(x)+e^{h(x)} h^{\prime}\left(e^xight) \cdot e^x \ & g^{\prime}(0)=h(1) e^{h(0)} h^{\prime}(0)+e^{h(0)} h^{\prime}(1) \ & =2+2=4\end{aligned}$

Suppose for a differentiable function $h, h(0)=0, h(1)=1$ and $h^{\prime}(0)=h^{\prime}(1)=2$. If…

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