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#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 42

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

Hint:  To solve this equation, we will convert them in log

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

Solution:

Taking log on both sides,

\begin{aligned} &\log (\cos x)^{y}=\log (\tan y)^{x} \\\\ &y \log (\cos x)=x \log (\tan x) \end{aligned}

$\frac{d}{d x}(y \log (\cos x))=\frac{d}{d x}(x \log (\tan x))$                        $\left[\begin{array}{l} \because \frac{d}{d x}(\log x)=\frac{1}{x} \\\\ \frac{d}{d x} \cos x=-\sin x \\\\ \frac{d}{d x}(\tan x)=\sec ^{2} x \end{array}\right]$

\begin{aligned} &\frac{d y}{d x} \cdot \log (\cos x)+y \frac{d}{d x}(\log (\cos x)) =\frac{d x}{d x} \cdot \log (\tan x)+x \cdot \frac{d}{d x}(\log (\tan x)) \end{aligned}

$\frac{d y}{d x} \log (\cos x)+y\left(\frac{-\sin x}{\cos x}\right)=\log (\tan x)+x \cdot \frac{\sec ^{2} y}{\tan y}-\frac{d y}{d x}$

$\frac{d y}{d x} \log (\cos x)-\tan x \cdot y=\log (\tan x)+x \cdot \frac{\frac{1}{\cos ^{2} y}}{\frac{\sin y}{\cos y}} \cdot \frac{d y}{d x}$

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

$\frac{d y}{d x}=\frac{\log (\tan x)+y \tan x}{\log (\cos x)-x \cos ec\; y \cdot \sec y}$