Derivative of tan ^{-1} sqrt{frac{1-x}{1+x}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $y = 	an ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.For $y$,substitute $x = \cos 2	heta$. Then $\sqrt{\frac{1-\cos 2	het

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(D) Let $y = 	an ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.For $y$,substitute $x = \cos 2	heta$. Then $\sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} = \sqrt{\frac{2\sin^2 	heta}{2\cos^2 	heta}} = 	an 	heta$.So,$y = 	an^{-1}(	an 	heta) = 	heta = \frac{1}{2} \cos^{-1} x$.For $z$,substitute $x = \cos 	heta$. Then $z = \cos^{-1}(\cos 3	heta) = 3	heta = 3 \cos^{-1} x$.Now,differentiate $y$ and $z$ with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} ight) = \frac{-1}{2\sqrt{1-x^2}}$.$\frac{dz}{dx} = 3 \left( \frac{-1}{\sqrt{1-x^2}} ight) = \frac{-3}{\sqrt{1-x^2}}$.Finally,$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-1/(2\sqrt{1-x^2})}{-3/\sqrt{1-x^2}} = \frac{1}{6}$.

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\cos ^{-1}\left(4 x^3-3 x\right)$ is

Similar Questions

$\frac{d}{dx} \left[ \sin^2 \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \right]$ equals

For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}$,the value of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

If $y = \tan^2 \left( \cos^{-1} \sqrt{\frac{1+x^2}{2}} \right)$,then $\frac{dy}{dx} = $

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

If $y = \tan^{-1} \left( \frac{x}{1 + \sqrt{1 - x^2}} \right) + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module