Differentiate

$\sin (2 \sin^{- 1} x)$

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Solution

$Let y = \sin (2 \sin^{- 1} x)$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} [\sin (2 \sin^{- 1} x)] = \cos (2 \sin^{- 1} x) \frac{d}{d x} (2 \sin^{- 1} x) [Using chain rule] = \cos (2 \sin^{- 1} x) \times 2 \frac{1}{\sqrt{1 - x^{2}}} = \frac{2 \cos (2 \sin^{- 1} x)}{\sqrt{1 - x^{2}}} So, \frac{d}{d x} [\sin (2 \sin^{- 1} x)] = \frac{2 \cos (2 \sin^{- 1} x)}{\sqrt{1 - x^{2}}}$

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	List I		List II
A.	Range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {cot}^{- 1} x$ is	1.	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
B.	Range of $f (x) = cot - 1 x + {tan}^{- 1} x + c o s e c^{- 1} x$ is	2.	$[\frac{π}{2}, \frac{3 π}{2}]$
C.	Range of $f (x) = {cot}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x$ is	3.	${0, π}$
D.	Range of $f (x) = {sec}^{- 1} x + c o s e c^{- 1} x + {sin}^{- 1} x$ is	4.	$(\frac{π}{2}, \frac{3 π}{2})$

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