Question

Derivative of $\sec^{-1} \left( \frac{1}{2x^2 + 1 } \right)$ w.r.t. $\sqrt{ 1 + 3x} $ at $x = - \frac{1}{3}$ is

• A. 0
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

Let $y = \sec^{-1} \left(\frac{1}{2x^{2}+1}\right) $ and $z=\sqrt{1+3x}$
$ \therefore \frac{dy}{dz} = \frac{dy/dx}{dz/dz}$
$ = \left(2x^{2} +1\right) . \frac{1}{\sqrt{\left(\frac{1}{2x^{2} +1}\right)^{2} -1} }$
$ . \frac{-4x}{ \left(2x^{2}+1\right)^{2}} / \frac{1}{2} \left(1+3x\right)^{-1/2}.3$
$ = \frac{-4x}{\sqrt{1-\left(2x^{2}+1\right)^{2}}} . \frac{2}{3} \sqrt{1-3x}$
At $x = \frac{1}{3}, \frac{dy}{dz} = 0 $