#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 56 maths

Answer:   $\frac{d y}{d x}=\frac{y \tan x+\log \cos y}{\log (\cos x)+x \tan y}$

Hint: To solve this equation we add log on both sides

Given: $(\cos x)^{y}=(\cos y)^{x}$

Solution:

$(\cos x)^{y}=(\cos y)^{x}$

Taking log on both sides,

$\log (\cos x)^{y}=\log (\cos y)^{x}$

$y \log (\cos x)=x \log (\cos y) \quad\left[\because \log m^{n}=n \log \mathrm{m}\right]$

$\frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}$

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

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

$-y \tan x+\log (\cos x) \frac{d y}{d x}=\frac{x(-\sin y)}{\cos y} \frac{d y}{d x}+\log \cos y$

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

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

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