frac{d}{dx} left( sin^{-1} left( frac{3+4x}{5 | vedclass

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

Question

(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} \right)$.Let $\cos \alpha = \frac{3}

Vedclass · Accepted Answer

$\frac{1}{1+x^2}$

Vedclass · Answer

(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} ight)$.Let $\cos \alpha = \frac{3}{5}$,then $\sin \alpha = \frac{4}{5}$.Also,let $\sin 	heta = \frac{x}{\sqrt{1+x^2}}$,then $\cos 	heta = \frac{1}{\sqrt{1+x^2}}$.Substituting these,we get $y = \sin^{-1} (\cos \alpha \cos 	heta + \sin \alpha \sin 	heta) = \sin^{-1} (\cos(\alpha - 	heta)) = \sin^{-1} (\sin(\frac{\pi}{2} - (\alpha - 	heta))) = \frac{\pi}{2} - \alpha + 	heta$.Since $	heta = 	an^{-1} x$,we have $y = \frac{\pi}{2} - \alpha + 	an^{-1} x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 0 - 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.

$\frac{d}{dx} \left( \sin^{-1} \left( \frac{3+4x}{5\sqrt{1+x^2}} \right) \right) =$

Similar Questions

If $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$,then $\frac{d y}{d x}=$

If $y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)$,then $\frac{d y}{d x}=$

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

The derivative of ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\frac{x}{2}$,where $x \in \left( {0, \frac{\pi }{2}} \right)$,is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module