If $y = \sin^{-1}(\sqrt{x})$,then $\frac{dy}{dx} = $
- A$\frac{2}{\sqrt{x}\sqrt{1-x}}$
- B$-\frac{2}{\sqrt{x}\sqrt{1-x}}$
- C$\frac{1}{2\sqrt{x}\sqrt{1-x}}$
- D$\frac{1}{\sqrt{1-x}}$
Explore More
Similar Questions
$\frac{d}{d x} \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$ is equal to
$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $
Medium
View SolutionIf $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$,then $\frac{dy}{dx} =$
If $y = \tan^{-1}(\sec x + \tan x)$,then $\frac{dy}{dx} = $
Find the derivative of $\cos^{-1} \sqrt{\frac{1 + x^2}{2}}$ with respect to $x$.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo