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 $x=\frac{1}{3},\frac{dy}{dz}=0$