Question

If $y=\left(1+x^2\right){\tan }^{-1}x-x$ then $\frac{dy}{dx}$ is

• A. $2x{\tan }^{-1}x$
• B. $x^2{\tan }^{-1}x$
• C. $\frac{{\tan }^{-1}x}{x}$
• D. $x{\tan }^{-1}x$

Answer 1

Answer: (A)

$y=\left(1+x^2\right){\tan }^{-1}x-x$
$\frac{dy}{dx}=\frac{\left(1+x^2\right)}{1+x^2}+{\tan }^{-1}x\cdot (2x)-1$
$=2x{\tan }^{-1}x$