#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 10

Answer:$\frac{y}{x}\left[\frac{x e^{(x-y)}-1}{y e^{(x-y)}-1}\right]$

Hint:

Use chain rule and quotient rule

Given:

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

Solution:

Differentiate the given equation w.r.t x

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

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

$\left(e^{x-y}\right) \times\left[\frac{d x}{d x}-\frac{d y}{d x}\right]=\frac{1}{\left(\frac{x}{y}\right)} \times\left[\frac{y \cdot \frac{d x}{d x}-x \cdot \frac{d y}{d x}}{y^{2}}\right]$                 $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x}\right]$

[Using quotient rule $\frac{d\left(\frac{u}{v}\right)}{d x}=\frac{v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}}{v^{2}}$    ]

$e^{x-y} \times\left[1-\frac{d y}{d x}\right]=\frac{y}{x} \times\left[\frac{y-x \frac{d y}{d x}}{y^{2}}\right]$

$e^{x-y}-e^{x-y} \frac{d y}{d x}=\frac{1}{x y}\left(y-x \frac{d y}{d x}\right)$

$e^{x-y}-e^{x-y} \frac{d y}{d x}=\frac{y}{x y}-\frac{x}{x y} \cdot \frac{d y}{d x}$

$\frac{x}{x y} \frac{d y}{d x}-e^{x-y} \frac{d y}{d x}=\frac{y}{x y}-e^{x-y}$

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

$\frac{d y}{d x}=\left[\frac{\frac{y}{x y}-e^{x-y}}{\frac{x}{x y}-e^{x-y}}\right]$

$\frac{d y}{d x}=\left[\frac{\frac{1}{x}-e^{x-y}}{\frac{1}{y}-e^{x-y}}\right]$

$=\frac{\left(\frac{1-x e^{x-y}}{x}\right)}{\left(\frac{1-y e^{x-y}}{y}\right)}$

$=\frac{y\left(1-x e^{x-y}\right)}{x\left(1-y e^{x-y}\right)}$

$=\frac{-y\left(x e^{x-y}-1\right)}{-x\left(y e^{x-y}-1\right)}$

$\frac{d y}{d x}=\frac{y}{x} \cdot\left(\frac{x e^{x-y}-1}{y e^{x-y}-1}\right)$

Hence $\frac{d y}{d x}=\frac{y}{x} \left[\frac{x e^{x-y}-1}{y e^{x-y}-1}\right]$ is the required differentiation