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 $\frac{dy}{dx}=0$