First slide

Derivatives

Question

If $y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1$ ,then $\frac{dy}{dx}$ is equal to

Easy

A

$\frac{- 1}{1 + x^{2}}$
B

$\frac{1}{1 + x^{2}}$
C

$\frac{- 2}{1 + x^{2}}$
D

$\frac{2}{1 + x^{2}}$

Solution

Substitute $x = \tan θ \Rightarrow θ = \tan^{- 1} x$ $y = \cos^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ})$
$\Rightarrow y = \cos^{- 1} (\sin 2 θ) [∵ \sin 2 θ = \frac{2 \tan θ}{1 + \tan^{2} θ}]$
$\Rightarrow y = \cos^{- 1} [\cos (\frac{π}{2} - 2 θ)]$ $[∵ \sin 2 θ = \cos (\frac{π}{2} - 2 θ)]$
$\Rightarrow y = \frac{π}{2} - 2 θ \Rightarrow y = \frac{π}{2} - 2 \tan^{- 1} x [∵ θ = \tan^{- 1} x]$
On differentiating both sides vv.r.t.x, we get
$\begin{array}{l} \frac{dy}{dx} = 0 - \frac{2}{1 + x^{2}} \\ \Rightarrow \frac{dy}{dx} = \frac{- 2}{1 + x^{2}} [∵ \frac{d}{dx} (\tan^{- 1} x) = \frac{1}{1 + x^{2}}] \end{array}$

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