First slide

Derivatives

Question

$\frac{d}{d x} \{\cos^{2} [\tan^{- 1} ({sincot}^{- 1} x)]\} =$

Moderate

A

$\frac{2}{{(x^{2} + 2)}^{2}}$
B

$\frac{2 x}{{(x^{2} + 2)}^{2}}$
C

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

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

Solution

$\frac{d}{d x} \{\cos^{2} [\tan^{- 1} ({sincot}^{- 1} x)]\} = \frac{d}{d x} \{\cos^{2} [T a n^{- 1} (\sin (\sin^{- 1} \frac{1}{\sqrt{1 + x^{2}}}))]\} = \frac{d}{d x} \{\cos^{2} [\tan^{- 1} \frac{1}{\sqrt{1 + x^{2}}}]\}$
$\frac{d}{d x} \{\frac{1 + \cos (2 \tan^{- 1} \frac{1}{\sqrt{1 + x^{2}}})}{2}\} = \frac{1}{2} \frac{d}{d x} \{1 + \frac{1 - \frac{1}{1 + x^{2}}}{1 + \frac{1}{1 + x^{2}}}\} = \frac{1}{2} \frac{d}{d x} \{1 + \frac{x^{2}}{2 + x^{2}}\} = \frac{1}{2} \frac{d}{d x} \{\frac{2 (1 + x^{2})}{2 + x^{2}}\}$
$= \frac{d}{d x} \{1 - \frac{1}{2 + x^{2}}\} = \frac{1}{{(2 + x^{2})}^{2}} (2 x) = \frac{2 x}{{(2 + x^{2})}^{2}} .$

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