#### Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 7

$\frac{\log x}{(1+\log x)^{2}}$

Hint:

Differentiate the function w.r.t x

Given:

$x^{y}=e^{x-y}$

Solution:

$x^{y}=e^{x-y}$

Taking log on both sides

$\log x^{y}=\log \left(e^{x-y}\right)$

$y \log x=(x-y) \log e \quad\left[\because \log m^{n}=n \log m\right]$

\begin{aligned} &y \log x=(x-y) \cdot 1\quad\quad\quad[\because \log e=1] \\\\ &y \log x=x-y \end{aligned}

\begin{aligned} &x=y+y \log x \\\\ &x=y(1+\log x) \\\\ &y=\frac{x}{1+\log x} \end{aligned}

Differentiating y w.r.t x then

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{1+\log x}\right)$

$=\frac{(1+\log x) \frac{d(x)}{d x}-x \frac{d}{d x}(1+\log x)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$

$=\frac{(1+\log x) \cdot 1-x\left\{\frac{d(1)}{d x}+\frac{d(\log x)}{d x}\right\}}{(1+\log x)^{2}}\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x), \frac{d(x)}{d x}=1\right]$

$=\frac{1+\log x-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}} \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x}(\operatorname{con} s \tan t)=0\right]$

$=\frac{1+\log x-x \cdot \frac{1}{x}}{(1+\log x)^{2}}=\frac{1+\log x-1}{(1+\log x)^{2}}$

$=\frac{\log x}{(1+\log x)^{2}}$