#### Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 11

The answer of the given question will be 2

Given:

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

Hint:

$\frac{d}{d x} u v=u \cdot \frac{d}{d x} \cdot v+. v \frac{d}{d x} u$

Solution:

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

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

\begin{aligned} &\Rightarrow \frac{d y}{d x}=f^{\prime}\left(e^{x}\right) \cdot e^{x} \cdot e^{f(x)}+f\left(e^{x}\right) \cdot e^{f(x)} \cdot f^{\prime}(x) \\\\ &\Rightarrow \frac{d y}{d x}=f^{\prime}\left(e^{0}\right) \cdot e^{0} \cdot e^{f(0)}+f\left(e^{0}\right) \cdot e^{f(0)} \cdot f^{\prime}(0) \end{aligned}

\begin{aligned} &=f^{\prime}(1) \cdot 1 \cdot e^{f(0)}+f(1) \cdot e^{f(0)} \cdot f^{\prime}(0) \\\\ &=2 . e^{0}+0 . e^{0} \cdot 2 \\\\ &=2.1+0 \\\\ &=2 \end{aligned}

So the answer of $\frac{d y}{d x}=2$