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  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 60 maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 60 maths

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Answer: \frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}

Hint: To solve this equation we use log on both side

Given: y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}

Solution:  \frac{d}{d x}(\log u)=\log (x \tan x)=\tan x-\log x

        \text { Let } \mathrm{u}=x^{\tan x}

        \frac{1}{u} \frac{d u}{d x}=\tan x \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x}(\tan x)

        \begin{aligned} &\frac{d u}{d x}=u\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \\\\ &\frac{d u}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \end{aligned}

        \begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(u+\sqrt{\frac{x^{2}+1}{2}}\right) \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d}{d x}\left(\sqrt{\frac{x^{2}+1}{2}}\right) \end{aligned}

        \frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x\right)  +\frac{1}{2 \sqrt{\frac{x^{2}+1}{2}}} \frac{d}{d x}\left(\frac{x^{2}+1}{2}\right)                \begin{aligned} &{\left[\because \frac{d}{d x} \sqrt{x}=\frac{1}{2 \sqrt{x}}\right]} \\ &{\left[\because \frac{d}{d x} x^{2}=2 x\right]} \end{aligned}

              =x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{1}{\sqrt{2\left(x^{2}+1\right)}} \cdot \frac{2 x}{2}

       \frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}

        

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