#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 20 maths

$\frac{d y}{d x}=-\frac{y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$

Hint:

Use product rule and chain rule

Given:

$x y \log (x+y)=1$

Solution:

$x y \log (x+y)=1$

Differentiate the given equation w.r.t x

$\frac{d}{d x}[x y \log (x+y)]=\frac{d(1)}{d x}$

$x y \cdot \frac{d}{d x}(\log (x+y))+\log (x+y) \frac{d(x y)}{d x}=0$                        $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$x y\left[\frac{d \log (x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}\right]+\log (x+y)\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]=0$           [Use product rule and chain rule]

$x y\left[\frac{1}{x+y} \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)\right]+\log (x+y)\left[x \frac{d y}{d x}+y\right]=0$

$x y\left[\frac{1}{x+y}+\frac{1}{x+y} \frac{d y}{d x}\right]+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$

$\frac{x y}{x+y}+\frac{x y}{x+y} \frac{d y}{d x}+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$

$\frac{d y}{d x}\left[\frac{x y}{x+y}+x \log (x+y)\right]=-\frac{x y}{x+y}-y \log (x+y)$

Also   $x y \log (x+y)=1$                         $\left[\log (x+y)=\frac{1}{x y}\right]$

Put $\log (x+y)=\frac{1}{x y}$   in the above equation

$\frac{d y}{d x}\left[\frac{x y}{x+y}+x \cdot \frac{1}{x y}\right]=-\frac{x y}{x+y}-y \cdot \frac{1}{x y}$

$\frac{d y}{d x}\left[\frac{x y}{x+y}+\frac{1}{y}\right]=-\left(\frac{x y}{x+y}+\frac{1}{x}\right)$

$\frac{d y}{d x}\left[\frac{x y^{2}+(x+y)}{y(x+y)}\right]=-\left(\frac{x^{2} y+(x+y)}{x(x+y)}\right)$

$\frac{d y}{d x}\left[\frac{\left(x y^{2}+(x+y)\right)}{y}\right]=-\left[\frac{\left(x^{2} y+(x+y)\right)}{x}\right]$

\begin{aligned} &\frac{d y}{d x}=\frac{-\left(x^{2} y+x+y\right)}{\left(x y^{2}+x+y\right)} \times \frac{y}{x} \\ &\frac{d y}{d x}=\frac{-y}{x} \cdot\left(\frac{x^{2} y+x+y}{x y^{2}+x+y}\right) \end{aligned}

Thus, proved