The derivative of sin ^{-1}left(3 x-4 x^3righ | vedclass

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

Question

(B) Let $y = \sin ^{-1}\left(3 x-4 x^3ight)$.We know that $\sin(3	heta) = 3\sin	heta - 4\sin^3	heta$.Let $x = \sin	heta$,then $	heta = \sin^{-1

Vedclass · Accepted Answer

$\frac{-3}{\sqrt{1-x^2}}$

Vedclass · Answer

(B) Let $y = \sin ^{-1}\left(3 x-4 x^3ight)$.We know that $\sin(3	heta) = 3\sin	heta - 4\sin^3	heta$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$.Since $\frac{1}{2} Then $3	heta$ lies in the range $(\frac{\pi}{2}, \frac{3\pi}{2})$.In this range,$\sin^{-1}(\sin(3	heta)) = \pi - 3	heta$.Thus,$y = \pi - 3\sin^{-1}x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\pi - 3\sin^{-1}x) = 0 - 3 	imes \frac{1}{\sqrt{1-x^2}} = \frac{-3}{\sqrt{1-x^2}}$.

The derivative of $\sin ^{-1}\left(3 x-4 x^3\right)$ with respect to $x$ for $\frac{1}{2} < x < 1$ is:

Similar Questions

If $f(x)=\cos ^{-1}\left[\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right]$,then $f^{\prime}(0.5)$ is equal to

The derivative of $\operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is

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

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module