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  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 16 maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 16 maths

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Answer: (\tan x)^{\frac{1}{x}} \cdot(x \operatorname{cosec} x \sec x)-\ln \left ( \tan \left ( x \right ).\frac{1}{x^{2}} \right )

Hint: Diff by (\tan x)^{\frac{1}{x}}

Given: (\tan x)^{\frac{1}{x}}

Solution:  Let y=(\tan x)^{\frac{1}{x}}

Taking log on both sides

        \begin{aligned} &\log y=\log (\tan x)^{\frac{1}{x}} \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log \tan x \end{aligned}

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{-1}{x^{2}} \ln (\tan x)+\frac{1}{x}\left(\frac{1}{\tan x} \sec ^{2} x\right) \\\\ &\left(\frac{1}{\tan x}=\cot x\right) \end{aligned}

So,

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y\left(\frac{-1}{x^{2}} \ln \left(\tan x+\frac{1}{x} \operatorname{cosec} x \cdot \sec x\right)\right. \\\\ &\frac{d y}{d x}=(\tan x)^{\frac{1}{x}}\left(\frac{-1}{x^{2}} \ln \tan x\right)+\frac{1}{x} \cos e c x \sec x \\\\ &\frac{d y}{d x}=(\tan x)^{\frac{1}{x}}\left[(x \operatorname{cosec} x \cdot \sec x)-\ln (\tan x) \cdot \frac{1}{x^{2}}\right] \end{aligned}       

Differentiate w.r.t x,

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