#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 27

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

Hint:

Use chain rule

Given:

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

Solution:

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

Differentiate w.r.t x

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

$\frac{d\left(e^{x}\right)}{d x}+\frac{d\left(e^{y}\right)}{d x}=\frac{d\left(e^{x+y}\right)}{d(x+y)} \times \frac{d(x+y)}{d x}$                        [Using chain rule]

$e^{x}+\frac{d\left(e^{y}\right)}{d y} \times \frac{d y}{d x}=e^{x+y} \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)$                    $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x}\right]$

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

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

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

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

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

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

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

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

Thus proved