#### Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 4 maths

The answer of the given question will be $\frac{1}{2}$.

Given:

If $f(1)=4 \text { and } f^{\prime}(1)=2$ , find the value of derivative $\log \left[f\left(e^{x}\right)\right]$ with respect to $x$ at the point $x=0$

Hint:

$\frac{d}{d x}(\log x)=\frac{1}{x} \& \frac{d}{d x}\left(e^{x}\right)=e^{x} \& \frac{d}{d x}[f(x)]=f^{\prime}(x)$

Solution:

\begin{aligned} &\frac{d u}{d x}=\frac{d}{d x} \log \left[f\left(e^{x}\right)\right] \\\\ &=\frac{1}{f\left(e^{x}\right)} \cdot \frac{d}{d x}\left[f\left(e^{x}\right)\right] \end{aligned}

\begin{aligned} &=\frac{1}{f\left(e^{x}\right)} \cdot f^{\prime}\left(e^{x}\right) \cdot \frac{d}{d x}\left(e^{x}\right) \\\\ &\Rightarrow \frac{d u}{d x}=\frac{f^{\prime}\left(e^{x}\right) \cdot e^{x}}{f\left(e^{x}\right)} \end{aligned}

\begin{aligned} &\Rightarrow \frac{d u}{d x}(\text { at } x=0)=\frac{f^{\prime}\left(e^{0}\right) \cdot e^{0}}{f\left(e^{0}\right)} \\\\ &=\frac{f^{\prime}(1) \cdot 1}{f(1)} \end{aligned}

\begin{aligned} &=\frac{2}{4} \\\\ &=\frac{1}{2} \end{aligned}

∴So, the answer will be $\frac{1}{2}$