Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

Given,

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

Now, differentiate with respect to $x$.

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& 2x\left({\tan }^{-1}x\right)+\left(1+x^2\right)\frac{1}{1+x^2}-1\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{u}\mathbf{\text{'}}\mathbf{v}\mathbf{}\mathbf{+}\mathbf{}\mathbf{v}\mathbf{\text{'}}\mathbf{u}\mathbf{,}\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{]}\\ & =& 2x\left({\tan }^{-1}x\right)+1-1\\ & =& 2x\left({\tan }^{-1}x\right)\end{array}$

Hence, the correct option is B.