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Please solve RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 17 maths textbook solution

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Answer:

The value of

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

Hint:

Using the chain rule of differentiation.

Given:

\tan ^{-1}\left(\frac{1-x}{1+x}\right)
Solution:  

y=\tan ^{-1}\left(\frac{1-x}{1+x}\right)

Using the chain rule of differentiation

\frac{d y}{d x}=\frac{1}{1+\left(\frac{1-x}{1+x}\right)^{2}} \cdot \frac{(1+x) \cdot(1-x)^{1}-(1+x)^{1} \cdot(1-x)}{(1+x)^{2}}

      \begin{aligned} &=\frac{(1+x)^{2}}{(1+x)^{2}+(1-x)^{2}} \cdot \frac{(1+x) \cdot(-1)-(1) \cdot(1-x)}{(1+x)^{2}} \\\\ &=\frac{-2}{(1+x)^{2}+(1-x)^{2}} \end{aligned}

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

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