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Mathematics Question on Continuity and differentiability

The derivative of cos⁡−1(1−x21+x2)\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) with respect to cot⁡−1(1−3x23x−x3)\cot^{-1} \left(\frac{1-3x^{2}}{3x-x^{3}}\right) is

A

12 \frac{1}{2}

B

11

C

−12 \frac{-1}{2}

D

−23 - \frac{2}{3}

Answer

−23 - \frac{2}{3}

Explanation

Solution

Let y=cos⁡−1(1−x21+x2)y = \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) ...(i)
and z=cot⁡−1(1−3x23x−x3)z = \cot^{-1} \left(\frac{1-3x^{2}}{3x-x^{3}}\right) ....(ii)
Putting x=tan⁡θx = \tan \theta in (i) and x=cot⁡ tx = \cot \, t in (ii), respectively
y=cos⁡−1(1−tan⁡2θ1+tan⁡2θ)y = \cos ^{-1}\left(\frac{1-\tan^{2} \theta}{1+\tan ^{2} \theta }\right)
=cos⁡−1(cos⁡2θ)=2tan⁡−1x= \cos ^{-1} \left(\cos 2\theta\right) =2 \tan^{-1} x
z=cot⁡−1(1−3cot⁡2t3cot⁡t−cot⁡3t)z = \cot^{-1}\left(\frac{1-3 \cot ^{2} t }{3 \cot t -\cot^{3} t }\right)
=cot⁡−1(cot⁡3t)=3cot⁡−1x=\cot ^{-1} \left(\cot 3t\right) =3 \cot ^{-1} x
dydx=21+x2\frac{dy}{dx}= \frac{2}{1+x^{2}} and dzdx=−31+x2\frac{dz}{dx} = \frac{-3}{1+x^{2}}
dydz=dydx.dxdz=(21+x2)(1+x2−3)=−23\frac{dy}{dz} = \frac{dy}{dx}. \frac{dx}{dz} = \left(\frac{2}{1+x^{2}}\right)\left(\frac{1+x^{2}}{-3}\right)= \frac{-2}{3}