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

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

Answers (1)

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}

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