Question

If $ y={{\sin }^{-1}}(3x-4{{x}^{3}})+{{\cos }^{-1}}(4{{x}^{3}}-3x) $ $ +{{\tan }^{-1}}(e), $ then $ \frac{dy}{dx} $ is equal to

• A. $ 5 $
• B. $ 0 $
• C. $ \frac{2}{\sqrt{1+{{x}^{2}}}} $
• D. $ \frac{2}{\sqrt{1-{{x}^{2}}}} $
• E. $ \frac{5}{\sqrt{1-{{x}^{2}}}} $

Answer 1

Answer: (B)

Given, $ y={{\sin }^{-1}}{{(3x-4x)}^{2}}+{{\cos }^{-1}}(4{{x}^{3}}-3x) $ $ +{{\tan }^{-1}}(E) $
$ \Rightarrow $ $ y=3{{\sin }^{-1}}x+3{{\cos }^{-1}}x+{{\tan }^{-1}}(E) $
$ \Rightarrow $ $ y=\frac{3\pi }{2}+{{\tan }^{-1}}(E) $
On differentiating w.r.t. $ x, $ we get $ \frac{dy}{dx}=0 $