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

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

Answers (1)

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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