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Find \(\large \frac {dy}{dx} \) in the following: $y = sin^{-1} \big( 2x \sqrt{1 -x^2} \big), -\;{ \frac{1}{\sqrt2} }\;< x < \;{\frac{1}{\sqrt2} }\\ $

$\begin{array}{1 1} \large\frac{dy}{dx}=\large\frac{-2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1+x^2}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1-x^3}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1-x^2}} \end{array} $

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Toolbox:
  • $\sin 2\theta=2\sin\theta\cos\theta$
  • $1-\sin^2\theta=\cos^2\theta$
Step 1:
Put $x=\sin\theta$
$y=\sin ^{-1}(2\sin \theta\sqrt{1-\sin^2\theta})$
We know that $\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=\cos\theta$
$y=\sin^{-1}(2\sin\theta\cos\theta)$
$y=\sin^{-1}(\sin 2\theta)$
$y=2\theta$
Step 2:
$y=2\sin^{-1}x$
We know that $\sin^{-1}x=\large\frac{1}{\sqrt{1-x^2}}$
$\large\frac{dy}{dx}=\large\frac{2}{\sqrt{1-x^2}}$
answered May 8, 2013 by sreemathi.v
 

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