#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 38

Answer:  $\frac{d y}{d x}=\frac{x(1+\log (x+y)-y x \log y}{y \log x+x-y(1+\log x+y)}$

Hint:  To solve this we add log on both sides

Given:   $x^{y}+y^{x}=(x+y)^{x+y}$

Solution:

Taking log on both sides,

\begin{aligned} &\log x^{y}+\log y^{x}=\log (x+y)^{x+y} \\\\ &y \log x+x \log y=(x+y)+\log (x+y) \end{aligned}                $\left[\because \frac{d}{d x}(u-v)=u . d v+v \cdot d u\right]$

$y \cdot \log x+x \log y^{\prime}+\frac{x}{y} y^{\prime}+\log y=\frac{(x+y)}{(x+y)}\left(1+y^{\prime}\right)+\log (x+y)\left(1+y^{\prime}\right)$

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

$y^{\prime}\left(\log x+\frac{x}{y}-(1+\log (x+y))=1+\log (x+y)-\left(\frac{y}{x}+\log y\right)\right.$

$y^{\prime}=\frac{x\left(1+\log (x+y)^{\prime-y}+x \log y\right)}{y \log x+x-y(1+\log (x+y))}$