#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 20 maths

Answer: $\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$

Hint:   $\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1} a-\tan ^{-1} b$
Given:  $\tan ^{-1}\left(\frac{1-x}{1+x}\right) \text { w.r.t } \sqrt{1-x^{2}}$

$-1

Explanation:

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

\begin{aligned} &u=\tan ^{-1}(1)-\tan ^{-1} x \\\\ &u=\frac{\pi}{4}-\tan ^{-1} x \\\\ &\frac{d u}{d x}=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} \end{aligned}

\begin{aligned} &v=\sqrt{1-x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\\\ &\frac{d v}{d x}=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned}

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\frac{1}{1+x^{2}}}{\frac{-x}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$

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Answer: $\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$

Hint:   $\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1} a-\tan ^{-1} b$
Given:  $\tan ^{-1}\left(\frac{1-x}{1+x}\right) \text { w.r.t } \sqrt{1-x^{2}}$

$-1

Explanation:

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

\begin{aligned} &u=\tan ^{-1}(1)-\tan ^{-1} x \\\\ &u=\frac{\pi}{4}-\tan ^{-1} x \\\\ &\frac{d u}{d x}=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} \end{aligned}

\begin{aligned} &v=\sqrt{1-x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\\\ &\frac{d v}{d x}=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned}

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\frac{1}{1+x^{2}}}{\frac{-x}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{x\left(1+x^{2}\right)}$