Question

Derivative of $ {{\sec }^{-1}}\left( \frac{1}{1-2{{x}^{2}}} \right) $ w.r.t. $ {{\sin }^{-1}}(3x-4{{x}^{3}}) $ is:

• A. $ \frac{1}{4} $
• B. $ \frac{3}{2} $
• C. 1
• D. $ \frac{2}{3} $
• E. $ -\frac{2}{3} $

Answer 1

Answer: (D)

$ u={{\sec }^{-1}}\left( \frac{1}{1-2{{x}^{2}}} \right),v={{\sin }^{-1}}(3x-4{{x}^{3}}) $ Let $ x=\sin \theta $ $ \Rightarrow $ $ u{{\sec }^{-1}}(\sec 2\theta ),v={{\sin }^{-1}}(\sin 3\theta ) $ $ \Rightarrow $ $ u=2\theta ,v=3\theta $ $ \Rightarrow $ $ u=2{{\sin }^{-1}}x,v=3{{\sin }^{-1}}x $ On differentiating w.r.t. $ \theta $ respectively $ \frac{du}{d\theta }=\frac{2}{\sqrt{1-{{x}^{2}}}},\frac{dv}{d\theta }=\frac{3}{\sqrt{1-{{x}^{2}}}} $ $ \therefore $ $ \frac{du}{d\theta }=\frac{\frac{\frac{du}{d\theta }}{dv}}{d\theta }=\frac{2}{3} $