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Question:

If $f(0)=f(1)=0, f^{\prime}(1)=2$ and $y=f\left(e^{x}\right) e^{f(x)}$, write the value of $\frac{d y}{d x}$ at $x=0$.

Solution:

Using the Chain Rule of Differentiation,

$\frac{d y}{d x}=u \cdot v^{\prime}+u^{\prime} \cdot v$

$=f\left(e^{x}\right) \cdot e^{f(x)} f^{\prime}(x)+f^{\prime}\left(e^{x}\right) e^{x} \cdot e^{f(x)}$

At $x=0$,

$\frac{d y}{d x}=f\left(e^{0}\right) \cdot e^{f(0)} f^{\prime}(0)+f^{\prime}\left(e^{0}\right) e^{0} \cdot e^{f(0)}$

$=f(1) \cdot e^{f(0)} f^{\prime}(0)+f^{\prime}(1) \cdot e^{f(0)}$

$=0 \cdot e^{0} f^{\prime}(0)+2 \cdot e^{0}$

$=0+2 \cdot 1$

$=2$