Question

If $y = (1+x^2) \tan^{-1}(x) - x$. 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)

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

Answer 2

The derivative of $(1 + x^2) \tan^{-1}(x)$ with respect to x can be found using the product rule and the chain rule

$\frac{d}{dx} \left[ (1 + x^2) \tan^{-1}(x) \right] = \frac{d}{dx} (1 + x^2) \tan^{-1}(x) + (1 + x^2) \frac{d}{dx} \tan^{-1}(x)$
The derivative of $(1 + x^2)$ with respect to x is 2x, and the derivative of $\tan^{-1}(x)$ with respect to x is $\frac{1}{1 + x^2}$
Therefore, we have:
$=(2x) \tan^{-1}(x) + \frac{1 + x^2}{1 + x^2}$ [Using the chain rule]
$=(2x) \tan^{-1}(x) + 1$
Now, let's differentiate the term -x:
$\frac{d}{dx}(-x) = -1$
Finally, we can add the derivatives of both terms:
$\frac{dy}{dx} = 2x \tan^{-1}(x) + 1 - 1$
Simplifying, we get:
$\frac{dy}{dx} = 2x \tan^{-1}(x)$
Therefore, the correct option is (A) $2x \tan^{-1}(x)$