If y=sqrt{frac{1-sin ^{-1} x}{1+sin ^{-1} x}} | vedclass

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

Question

(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$Differentiating with respect to $x$,we get$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$Differentiating with respect to $x$,we get$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\right)$$= \frac{1}{2 \sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}} \cdot \frac{d}{dx}\left(\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}\right)$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{d}{dx}(1-\sin ^{-1} x) - (1-\sin ^{-1} x) \cdot \frac{d}{dx}(1+\sin ^{-1} x)}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{-1}{\sqrt{1-x^2}} - (1-\sin ^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \left[ \frac{-1-\sin ^{-1} x - 1 + \sin ^{-1} x}{(1+\sin ^{-1} x)^2} \right]$$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \frac{-2}{(1+\sin ^{-1} x)^2}$$= \frac{-\sqrt{1+\sin ^{-1} x}}{\sqrt{1-\sin ^{-1} x} \cdot \sqrt{1-x^2} \cdot (1+\sin ^{-1} x)^2}$At $x=0$,$\sin^{-1}(0) = 0$:$\left(\frac{dy}{dx}\right)_{x=0} = \frac{-\sqrt{1+0}}{\sqrt{1-0} \cdot \sqrt{1-0} \cdot (1+0)^2} = \frac{-1}{1 \cdot 1 \cdot 1} = -1$

If $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$,then $\left(\frac{dy}{dx}\right)$ at $x=0$ is

Similar Questions

If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

If $y = \cos^{-1} \left( \frac{1-4^x}{1+4^x} \right)$, then find $\frac{dy}{dx}$ at $x = 1$.

The differential coefficient of $\sin^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to $\sqrt{x}$ is:

If $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$,then $\frac{d y}{d x}$ is

If $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module