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# Solution for Differentiate Cos − 1 { 2 X √ 1 − X 2 } , 1 √ 2 < X < 1 ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

Differentiate $\cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}, \frac{1}{\sqrt{2}} < x < 1$ ?

#### Solution

$\text{ Let, y } = \cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}$

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

$y = \cos^{- 1} \left\{ 2\cos\theta\sqrt{1 - \cos^2 \theta} \right\}$

$y = \cos^{- 1} \left\{ 2\cos\theta \sin\theta \right\}$

$y = \cos^{- 1} \left\{ \sin2\theta \right\} \left[ Since, \sin2\theta = 2\sin\theta\cos\theta \right]$

$y = \cos^{- 1} \left[ \cos\left( \frac{\pi}{2} - 2\theta \right) \right] . . . \left( i \right)$

$\text{Now,}$

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

$\Rightarrow \frac{1}{\sqrt{2}} < \cos\theta < 1$

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

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

$\Rightarrow 0 > - 2\theta > - \frac{\pi}{2}$

$\Rightarrow \frac{\pi}{2} > \left( \frac{\pi}{2} - 2\theta \right) > 0$

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

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

$y = \frac{\pi}{2} - 2 \cos^{- 1} x \left[ Since, x = \cos\theta \right]$

$\text{ differentiating it with respect to x },$

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\pi}{2} \right) - 2\frac{d}{dx}\left( \cos^{- 1} x \right)$

$\frac{d y}{d x} = 0 - 2\left( \frac{- 1}{\sqrt{1 - x^2}} \right)$

$\frac{d y}{d x} = \frac{2}{\sqrt{1 - x^2}}$

Is there an error in this question or solution?

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Solution Differentiate Cos − 1 { 2 X √ 1 − X 2 } , 1 √ 2 < X < 1 ? Concept: Simple Problems on Applications of Derivatives.
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