Question:

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

Updated On: Jul 6, 2022
  • 0
  • $\frac{1}{2}$
  • $\frac{1}{3}$
  • $\frac{1}{6}$
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The Correct Option is A

Solution and Explanation

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 $
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