Question

The derivative of $ \text{cose}{{\text{c}}^{-1}},\,\,\,\,\left( \frac{1}{2x\,\sqrt{1-{{x}^{2}}}} \right) $ with respect to $ \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)

Let $ u=\text{cose}{{\text{c}}^{-1}}\left( \frac{1}{2x\sqrt{1-{{x}^{2}}}} \right) $
and $ v=\sqrt{1-{{x}^{2}}} $
Put $ x=\sin \theta , $
Then, $ u=\text{cose}{{\text{c}}^{-1}}\left\{ \frac{1}{2\sin \theta .\cos \theta } \right\}=\text{cose}{{\text{c}}^{-1}}\,(\text{cosec}\,\text{ 2}\theta \text{)} $
$ u=2\theta =2{{\sin }^{-1}}x $ ..(i)
and $ v=\sqrt{1-{{\sin }^{2}}\theta }\,=\,\cos \theta =\sqrt{1-{{x}^{2}}} $ ..(ii)
Now, $ \frac{du}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}} $
and $ \frac{dv}{dx}=\frac{-2x}{2\sqrt{1-{{x}^{2}}}}=\frac{-x}{\sqrt{1-{{x}^{2}}}} $ 0
$ \Rightarrow $ $ \frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{2/\sqrt{1-{{x}^{2}}}}{-x/\sqrt{1-{{x}^{2}}}}=\frac{-2}{x} $