Question

The derivative of $cose{c}^{-1}\left(\frac{1}{2x\sqrt{1-x^2}}\right)w.r.t\sqrt{1-x^2}$ is

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{2}{x}$
• C. $-\frac{2}{x}$
• D. $-\frac{1}{\sqrt{1-x^2}}$

Answer 1

Answer: (C)

Put $x=\sin \theta$
Let $y=cose{c}^{-1}\left(\frac{1}{2x\sqrt{1-x^2}}\right)$
$=cose{c}^{-1}\left(\frac{1}{2\sin \theta \cos \theta }\right)=cose{c}^{-1}$
$\left(cosec2\theta \right)=2\theta$
$\therefore \frac{dy}{dz}=\frac{dy/d\theta }{dz/d\theta }=\frac{2}{-\sin \theta }=-\frac{2}{x}$
$z=\sqrt{1-x^2}=\sqrt{1-{\sin }^2\theta }=\cos \theta$