Question

The derivative of $cosec^{-1} \left(\frac{1}{2x\sqrt{1-x^{2}}}\right) w.r.t \sqrt{1-x^{2}}$ 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 = cosec^{-1} \left(\frac{1}{2x \sqrt{1-x^{2}}}\right)$
$ = cosec ^{-1} \left(\frac{1}{2\sin \theta \cos\theta}\right) = cosec^{-1} $
$\left(cosec 2 \theta\right) = 2 \theta$
$ \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 $