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Differentiate the following w.r.t x:$\tan^{-1}\bigg(\sqrt{\large\frac{1-\cos x}{1+\cos x}}\bigg),$$\large\frac{-\pi}{4}<$$x<\large\frac{\pi}{4}$.

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Toolbox:
  • $1-\cos x=2\sin^2\large\frac{x}{2}$
  • $1+\cos x=2\cos^2\large\frac{x}{2}$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
$y=\tan^{-1}\bigg(\sqrt{\large\frac{1-\cos x}{1+\cos x}}\bigg)$
But $1-\cos x=2\sin^2\large\frac{x}{2}$
$1+\cos x=2\cos^2\large\frac{x}{2}$
$y=\tan^{-1}\bigg(\sqrt{\large\frac{2\sin^2\large\frac{x}{2}}{2\cos^2\Large\frac{x}{2}}\bigg)}$
$y=\tan^{-1}\bigg(\large\frac{2\sin \Large\frac{x}{2}}{2\cos \Large\frac{x}{2}}\bigg)$
$y=\tan^{-1}(\tan\large\frac{x}{2})$
$y=\large\frac{x}{2}$
$\large\frac{dy}{dx}=\frac{1}{2}$
answered Jun 28, 2013 by sreemathi.v
 

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