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
$\Rightarrow$ $\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}}=\frac{-2}{x}$