Question

If $y=\left(1+x^2\right) \tan ^{-1} x-x$, then $\frac{d y}{d x}$ is equal to

(A) $\tan ^{-1} x$

(B) $2 x \tan ^{-1} x$

(C) $2 x \tan ^{-1} x-1$

(D) $\frac{2 x}{\tan ^{-1} x}$

Answer 1

$\text { } \begin{array}{l} y=\left(1+x^2\right) \tan ^{-1} x-x \\\Rightarrow \quad \frac{d y}{d x}=\left(1+x^2\right) \cdot \frac{1}{\left(1+x^2\right)}+\tan ^{-1} x \cdot(2 x)-1 \\=2 x \tan ^{-1} x\end{array}$

So, The correct option will be (B).