Question

The derivative of $se{c}^{-1} \left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is:

Answer 1

Correct Answer: 4

Given: The function, $se{c}^{-1} \left(\frac{1}{2x^2-1}\right)$
We have to find it's derivative with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$
Let $y = se{c}^{-1} \left(\frac{1}{2x^2-1}\right)$= cos^-1(2x² - 1)       [Since sec^-1$\left(\frac{1}{\theta }\right)$ = cos^-1θ]
Substituting x = cos θ ,
we get y = cos^-1(2cos²θ - 1)= cos^-1(cos²θ ) [Using trigonometric formula of cosine function ]
= 2θ
= 2 cos¹
Differentiating w.r.t. x, we get
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$     [Using standard derivatives-4 ]
Let $z=\sqrt{1-x^2}$    
Differentiating w.r.t. x, we get
$$\begin{array}{l}\frac{dz}{dx}=\frac{-x}{\sqrt{1-x^2}}\\ \therefore \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{\frac{-2}{\sqrt{1-x^2}}}{\frac{-x}{\sqrt{1-x^2}}}=\frac{2}{x}\\ {\left(\frac{dy}{dz}\right)}_{atx=\frac{1}{2}}=\frac{2}{\left(\frac{1}{2}\right)}=4\end{array}$$
Hence, the answer is 4.00