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Mathematics Question on Differentiability

The derivative of tan−1[Sinx1+Cosx]tan ^{-1}[\frac{Sin x}{1+ Cos x}] with respect to tan−1[Cosx1+Sinx]tan^{-1}[\frac{Cos x}{1+Sin x}] is

A

2

B

-1

C

0

D

-2

Answer

-1

Explanation

Solution

Let
u=tan⁡−1(sin⁡x1+cos⁡x),u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right),
v=tan⁡−1(cos⁡x1+sin⁡x)v=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)
u=tan⁡−1(2sin⁡x/2⋅cos⁡x/21+2cos⁡2x/2−1)u =\tan ^{-1}\left(\frac{2 \sin x / 2 \cdot \cos x / 2}{1+2 \cos ^{2} x / 2-1}\right)
=tan⁡−1(tan⁡x/2)=x/2=\tan ^{-1}(\tan x / 2)=x / 2
v=tan⁡−1(cos⁡xcos⁡2x2+sin⁡2x2+2sin⁡x2⋅cos⁡x2)v=\tan ^{-1}\left(\frac{\cos x}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)
=tan⁡−1((cos⁡2x2−sin⁡2x2)(cos⁡x2+sin⁡x2)2)=\tan ^{-1}\left(\frac{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \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⁡−1(π4−π2)]=(π4−x2)=\left[\tan ^{-1}\left(\tan ^{-1}\left(\frac{\pi}{4}-\frac{\pi}{2}\right)\right]=\left(\frac{\pi}{4}-\frac{x}{2}\right)\right.
Now, dudx=12\frac{du}{dx}=\frac{1}{2} and dvdx=12\frac{d v}{d x}=\frac{1}{2}
⇒dudv=dudx×dxdv\Rightarrow \frac{d u}{d v}=\frac{d u}{d x} \times \frac{d x}{d v}
=(12)×(−2)=−1=\left(\frac{1}{2}\right) \times(-2)=-1