Find frac{dy}{dx},if y = sin^{-1}x + sin^{-1} | vedclass

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

Question

(A) Given $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$. Since $-1 \le x \le 1$,we have $-\frac{\pi}{2} \le

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$. Since $-1 \le x \le 1$,we have $-\frac{\pi}{2} \le 	heta \le \frac{\pi}{2}$.Then $\sqrt{1-x^2} = \sqrt{1-\sin^2	heta} = \sqrt{\cos^2	heta} = |\cos	heta|$.Since $-\frac{\pi}{2} \le 	heta \le \frac{\pi}{2}$,$\cos	heta \ge 0$,so $|\cos	heta| = \cos	heta$.Thus,$y = \sin^{-1}(\sin	heta) + \sin^{-1}(\cos	heta) = 	heta + \sin^{-1}(\sin(\frac{\pi}{2} - 	heta))$.For $x \in [0, 1]$,$0 \le 	heta \le \frac{\pi}{2}$,so $y = 	heta + (\frac{\pi}{2} - 	heta) = \frac{\pi}{2}$.For $x \in [-1, 0]$,$-\frac{\pi}{2} \le 	heta $y = 	heta + \sin^{-1}(\sin(\frac{\pi}{2} + 	heta))$. Since $\frac{\pi}{2} + 	heta \in [0, \pi]$,we use the identity $\sin^{-1}(\sin A) = \pi - A$ for $A \in [\frac{\pi}{2}, \pi]$.$y = 	heta + \pi - (\frac{\pi}{2} + 	heta) = \frac{\pi}{2}$.In both cases,$y = \frac{\pi}{2}$.Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

Find $\frac{dy}{dx}$,if $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$,where $-1 \le x \le 1$.

Similar Questions

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

Suppose $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$,where $|x| < \frac{1}{\sqrt{3}}$,then one of the values of $y$ is

$\sin \left\{ {{\tan }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \right\}$ is equal to

Evaluate $\sin(3 \sin^{-1}(1/5))$.

$\sum_{i=0}^2 \cot ^{-1}\{-(i+1)\}=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module