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If Y = Sin − 1 ( 2 X 1 + X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , 0 < X < 1 , Prove that D Y D X = 4 1 + X 2 ? - CBSE (Arts) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

Question

If  $y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), 0 < x < 1,$ prove that  $\frac{dy}{dx} = \frac{4}{1 + x^2}$ ?

Solution

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

$\Rightarrow y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)$

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

$\therefore y = \sin^{- 1} \left( \frac{2 \tan\theta}{1 + \tan^2 \theta} \right) + \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)$

$\Rightarrow y = \sin^{- 1} \left( \sin 2\theta \right) + \cos^{- 1} \left( \cos 2\theta \right) . . . \left( i \right)$

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

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

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

$\Rightarrow 0 < 2\theta < \frac{\pi}{2}$

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

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

$\Rightarrow y = 4\theta$

$\Rightarrow y = 4 \tan^{- 1} x ...........\left[ \text{ Since}, x = \tan\theta \right]$

Differentiate it with respect to x,

$\therefore \frac{d y}{d x} = \frac{4}{1 + x^2}$

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Solution If Y = Sin − 1 ( 2 X 1 + X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , 0 < X < 1 , Prove that D Y D X = 4 1 + X 2 ? Concept: Simple Problems on Applications of Derivatives.
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