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# Solution for Differentiate Tan − 1 ( X √ 1 − X 2 ) with Respect to - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Questions

Differentiate $\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$ with respect to $\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right), \text { if } - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ ?

Differentiate $\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$ with respect to $\sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)$ .

#### Solution

$\text { Let, u } = \tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$

$\text { Put x } = \sin\theta$

$\Rightarrow \theta = \sin^{- 1} x$

$\Rightarrow u = \tan^{- 1} \left( \frac{\sin\theta}{\sqrt{1 - \sin^2 \theta}} \right)$

$\Rightarrow u = \tan^{- 1} \left( \frac{\sin\theta}{\cos\theta} \right)$

$\Rightarrow u = \tan^{- 1} \left( \tan\theta \right) . . . \left( i \right)$

$\text { And }$

$\text { Let, v } = \sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)$

$v = \sin^{- 1} \left( 2\sin\theta\sqrt{1 - \sin^2 \theta} \right)$

$v = \sin^{- 1} \left( 2 \sin\theta\cos\theta \right)$

$v = \sin^{- 1} \left( \sin2\theta \right) . . . \left( ii \right)$

$\text { Here,}$

$- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

$\Rightarrow - \frac{1}{\sqrt{2}} < \sin\theta < \frac{1}{\sqrt{2}}$

$\Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}$

$\text { So, from equation } \left( i \right),$

$u = \theta \left[ \text { Since,} \tan^{- 1} \left( \tan\theta \right) = \theta, \text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]$

$\Rightarrow u = \sin^{- 1} x$

Differentiating it with respect to x,

$\frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iii \right)$

$\text{ from equation } \left( ii \right),$

$v = 2\theta \left[ \text { Since}, \sin^{- 1} \left( \sin\theta \right) = \theta, \text { if } \theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]$

$\Rightarrow v = 2 \sin^{- 1} x$

Differentiating it with respect to x,

$\frac{dv}{dx} = \frac{2}{\sqrt{1 - x^2}} . . . \left( iv \right)$

$\text { Dividing equation } \left( iii \right) \text {by}\left( iv \right),$

$\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \left( \frac{1}{\sqrt{1 - x^2}} \right)\left( \frac{\sqrt{1 - x^2}}{2} \right)$

$\therefore \frac{du}{dv} = \frac{1}{2}$

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Solution Differentiate Tan − 1 ( X √ 1 − X 2 ) with Respect to Concept: Simple Problems on Applications of Derivatives.
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