Differentiation of tan ^{-1}left(frac{x}{sqrt | vedclass

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

Question

(D) Let $u = 	an ^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight)$ and $v = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,where $	heta = \s

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) Let $u = 	an ^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight)$ and $v = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,where $	heta = \sin ^{-1} x$.Then $u = 	an ^{-1}\left(\frac{\sin 	heta}{\sqrt{1-\sin ^2 	heta}}ight) = 	an ^{-1}\left(\frac{\sin 	heta}{\cos 	heta}ight) = 	an ^{-1}(	an 	heta) = 	heta$.And $v = \sin ^{-1}(3 \sin 	heta - 4 \sin ^3 	heta) = \sin ^{-1}(\sin 3 	heta) = 3 	heta$.We need to find $\frac{du}{dv}$.Since $u = 	heta$ and $v = 3 	heta$,we have $\frac{du}{d	heta} = 1$ and $\frac{dv}{d	heta} = 3$.Therefore,$\frac{du}{dv} = \frac{du/d	heta}{dv/d	heta} = \frac{1}{3}$.

Differentiation of $\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is $....$

Similar Questions

$\frac{d}{dx}\left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) = $

$\frac{d}{dx} \left\{ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right\} = $

The differential coefficient of ${\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ with respect to $\sqrt {1 - {x^2}} $ at $x = \frac{1}{2}$ is:

Find the derivative of $\cos^{-1} \sqrt{\frac{1 + x^2}{2}}$ with respect to $x$.

If $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$,then $\frac{dy}{dz} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module