First slide

Derivatives

Question

$If y = \sin^{- 1} (\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}) then \frac{d y}{d x} =$

Moderate

A

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

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

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

None

Solution

$\begin{array}{l} put x = \cos θ \\ 1 + \cos θ = 2 \cos^{2} \frac{θ}{2}; 1 - \cos θ = 2 \sin^{2} \frac{θ}{2} \\ y = \sin^{- 1} [\frac{1}{\sqrt{2}} (\cos \frac{θ}{2} + \sin \frac{θ}{2})] \\ = \sin^{- 1} \sin (\frac{θ}{2} + \frac{π}{4}) = \frac{θ}{2} + \frac{π}{4} \\ y = \frac{1}{2} \cos^{- 1} x + \frac{π}{4} \Rightarrow \frac{d y}{d x} = \frac{- 1}{2 \sqrt{1 - x^{2}}} \end{array}$

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