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  • Differentiate each of the following w.r.t. x tan inverse root 1 minus cos x by 1 plus cos x

Differentiate each of the following w.r.t. x
\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}

 

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We have \tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}

\begin{aligned} &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}\right)}\\ &\text { We know, }\\ &\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{\sin ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}-\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}}{\sin ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}-\sin ^{2} \frac{\mathrm{x}}{2}}\right)}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{2 \sin ^{2} \frac{\mathrm{x}}{2}}{2 \cos ^{2} \frac{\mathrm{x}}{2}}\right)}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\tan ^{2} \frac{\mathrm{x}}{2}\right)}\\ &y=\tan ^{-1}\left(\tan \frac{x}{2}\right) \end{aligned}

\begin{aligned} &\text { As the interval is }\\ &-\frac{\pi}{4}<x<\frac{\pi}{4}\\ &=\left\{\begin{array}{r} \tan ^{-1}\left(\tan \frac{x}{2}\right),-\frac{\pi}{4}<x<0 \\ \tan ^{-1}\left(\tan \frac{x}{2}\right), 0 \leq x<\frac{\pi}{4} \end{array}\right.\\ &=\left\{\begin{array}{c} -\frac{\mathrm{x}}{2},-\frac{\mathrm{\pi}}{4}<\mathrm{x}<0 \\ \frac{\mathrm{x}}{2}, 0 \leq \mathrm{x}<\frac{\pi}{4} \end{array}\right. \end{aligned}

Differentiate w.r.t. x

\frac{d y}{d x}=\left\{\begin{array}{c} -\frac{1}{2},-\frac{\pi}{4}<x<0 \\ \frac{1}{2}, 0 \leq x<\frac{\pi}{4} \end{array}\right.

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infoexpert22

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