#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 36 maths

Answer: $\frac{d y}{d x}=-\left\{\frac{\left.x^{x}(1+\log x)+y^{x} \log y\right)}{x \cdot y^{(x-1)}}\right\}$

Hint:  To solve this equation we denote both term as u and v

Given: $x^{x}+y^{x}=1$

Solution:

$\frac{d u}{d x}+\frac{d v}{d x}=0$

$u=x^{x}$

Taking log on both sides,

$\log u=x \log x$

on diff. we get

\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \frac{d}{d x}(\log x)+\log x \cdot \frac{d x}{d x} \\\\ &\frac{d u}{d x}=u\left(x \cdot \frac{1}{x}+\log x\right) \end{aligned}

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

Taking log both side

$\log v=x \log y$

on diff. we get

$\frac{1}{v} \cdot \frac{d v}{d x}=x \cdot \frac{d}{d x}(\log y)+\log y \cdot \frac{d x}{d x}$

\begin{aligned} &\frac{d v}{d x}=v\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right] \\\\ &\frac{d v}{d x}=y^{x} \cdot\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right] \end{aligned}

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