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Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 17 maths textbook solution

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Answer: \frac{d y}{d x}=\left[\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}\right] \cdot \tan ^{-1} x

Hint: Diff by x^{\tan ^{-1} x}

Given: x^{\tan ^{-1} x}

Solution:  Let y=x^{\tan ^{-1} x}

Taking log on both sides

        \begin{aligned} &\log y=\log x^{\tan ^{-1} x} \\\\ &\frac{1}{y} \frac{d y}{d x}=\tan ^{-1} x \cdot \frac{1}{x}+\log x \cdot \frac{1}{1+x^{2}} \end{aligned}

        \frac{1}{y} \frac{d y}{d x}=\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}

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


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