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Find \( \large\frac {dy}{dx} \) in the following: $ y = cos^{-1} \left(\large\frac{2x}{1 + x^2}\right), -1 < x < 1 $

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

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