Question

If $s={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ and $t=\sqrt{1-x^2}$, then $\frac{ds}{dt}$ at $x=\frac{1}{2}$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $4$
• E. $-4$

Answer 1

The correct option is D $4$

Given, $s={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

$\Rightarrow s={\cos }^{-1}(2x^2-1)$

$[\because {\cos }^{-1}x={\sec }^{-1}\frac{1}{x}]$

$\Rightarrow s=2{\cos }^{-1}x$

$[\because 2{\cos }^{-1}x={\cos }^{-1}(2x^2-1)]$

On differentiating w.r.t $x$ we get

$\frac{ds}{dx}=2.\frac{-1}{\sqrt{1-x^2}}=\frac{-2}{\sqrt{1-x^2}}$

also $t=\sqrt{1-x^2}$

On differentiating again

$\frac{dt}{dx}=\frac{1}{2\sqrt{1-x^2}}(-2x)=\frac{-x}{\sqrt{1-x^2}}$

Now, $\frac{ds}{dx}=\frac{ds}{dx}\times \frac{dx}{dt}$

$=\frac{-2}{\sqrt{1-x^2}}\times \frac{\sqrt{1-x^2}}{-x}=\frac{2}{x}$

$\therefore {\left(\frac{ds}{dt}\right)}_{x=\frac{1}{2}}=\frac{2}{1/2}=4$