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# Solution for Differentiate $\Sin^{- 1} \Left( \Frac{2x}{1 + X^2} \Right)$ With Respect to $\Cos^{- 1} \Left( \Frac{1 - X^2}{1 + X^2} \Right), \Text { If } 0 < X < 1$ ? - CBSE (Science) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

Differentiate $\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)$ with respect to $\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { if } 0 < x < 1$ ?

#### Solution

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

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

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

$\Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)$

$\text { Let v } = \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)$

$\Rightarrow v = \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)$

$\Rightarrow v = \cos^{- 1} \left( \cos2\theta \right) . . . \left( ii \right)$

$\text { Here }, 0 < x < 1$

$\Rightarrow 0 < \tan\theta < 1$

$\Rightarrow 0 < \theta < \frac{\pi}{4}$

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

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

$\Rightarrow u = 2 \tan^{- 1} x \left[ \text { Since } , x = \tan\theta \right]$

Differentiating it with respect to x,

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

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

$v = 2\theta \left[ \text { Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, if \theta \in \left[ 0, \pi \right] \right]$

$\Rightarrow v = 2 \tan^{- 1} x \left[ \text { Since}, x = \tan\theta \right]$

Differentiating it with respect to x,

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

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

$\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2}$

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

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Solution Differentiate $\Sin^{- 1} \Left( \Frac{2x}{1 + X^2} \Right)$ With Respect to $\Cos^{- 1} \Left( \Frac{1 - X^2}{1 + X^2} \Right), \Text { If } 0 < X < 1$ ? Concept: Simple Problems on Applications of Derivatives.
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