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Find \(\large \frac{dy}{dx} \), if \( y = \sin^{-1}x + \sin^{-1} \: \sqrt {1-x^2} , -1 \leq x \leq 1 \)

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Toolbox:
  • $\large\frac{d}{dx}$$(constant)=0$
  • $\sin^2\theta+\cos^2\theta=1$
Step 1:
Put $x=\sin \theta$
$\theta=\sin^{-1}x$
$y=\theta+\sin^{-1}\sqrt{1-\sin^2\theta}$
$\;\;=\theta+\sin^{-1}\sqrt{\cos^2\theta}$
$\;\;=\theta+\sin^{-1}(\cos\theta)$
$\;\;=\theta+\sin^{-1}\sin(\large\frac{\pi}{2}$$-\theta)$
$\;\;=\theta+\large\frac{\pi}{2}$$-\theta$
Step 2:
$y=\large\frac{\pi}{2}$
$\large\frac{dy}{dx}$$=0$
answered May 14, 2013 by sreemathi.v
 

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