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Mathematics Question on Differentiability

If y=tan⁡−1(cos⁡x1+sin⁡x),y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right), then dydx\frac{dy}{dx} is equal to

A

12\frac{1}{2}

B

22

C

−2-2

D

−12-\frac{1}{2}

Answer

−12-\frac{1}{2}

Explanation

Solution

Given that, y=tan⁡−1(cos⁡x1+sin⁡x)y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right)
=tan⁡−1(cos⁡2x2−sin⁡2x2sin⁡2x2+cos⁡2x2+2sin⁡x2.cos⁡x2)={{\tan }^{-1}}\left( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)
=tan⁡−1((cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)(cos⁡x2+sin⁡x2)2)={{\tan }^{-1}}\left( \frac{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}} \right)
=tan⁡−1(cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2)={{\tan }^{-1}}\left( \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right)
=tan⁡−1(1−tan⁡x21+tan⁡x2)={{\tan }^{-1}}\left( \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \right)
=tan⁡−1(tan⁡(π4−x2))={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}-\frac{x}{2} \right) \right)
=π4−x2=\frac{\pi }{4}-\frac{x}{2}
∴\therefore y=π4−x2y=\frac{\pi }{4}-\frac{x}{2}
Differentiating it w.r.t., xx we get dydx=−12\frac{dy}{dx}=-\frac{1}{2}