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

The Correct Option is A

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) \(2x \tan^{-1}(x)\)