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 $y=\tan ^{-1}\left[\frac{1}{1+x+x^2}\right]+\tan ^{-1}\left[\frac{1}{x^2+3 x+3}\right], x>0$,then $\frac{d y}{d x}=$

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ has the value

If $y = \tan^{-1}\left(\frac{a \cos x - b \sin x}{b \cos x + a \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

If $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right)$,then $f'(-\frac{1}{2})$ equals

If $f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$,then $f'(e) = \_\_\_\_$

Vedclass Test Series

Exam Paper Generator

Online Exam Module