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) \(2x \tan^{-1}(x)\)

Answer 2

We are given: \[ y = (1 + x^2) \tan^{-1}(x) - x \] Differentiate using product rule and chain rule: Let \( u = (1 + x^2),\ v = \tan^{-1}(x) \) Then, \[ \frac{dy}{dx} = \frac{d}{dx}[(1 + x^2)\tan^{-1}(x)] - \frac{d}{dx}[x] \] Apply product rule: \[ = (1 + x^2) \cdot \frac{d}{dx}[\tan^{-1}(x)] + \tan^{-1}(x) \cdot \frac{d}{dx}[1 + x^2] - 1 \] \[ = (1 + x^2) \cdot \frac{1}{1 + x^2} + \tan^{-1}(x) \cdot 2x - 1 \] \[ = 1 + 2x \tan^{-1}(x) - 1 \] \[ = \boxed{2x \tan^{-1}(x)} \] So, \(\frac{dy}{dx} = 2x \tan^{-1}(x)\) 

Answer 3

Given \(y = (1+x^2) \tan^{-1}(x) - x\), we need to find \(\frac{dy}{dx}\).

We will use the product rule for differentiation, which states that if y = uv, then \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\).

\(\frac{dy}{dx} = \frac{d}{dx}[(1+x^2) \tan^{-1}(x) - x]\)

\(\frac{dy}{dx} = \frac{d}{dx}[(1+x^2) \tan^{-1}(x)] - \frac{d}{dx}[x]\)

Let u = (1 + x²) and v = tan^-1(x).

\(\frac{du}{dx} = 2x\)

\(\frac{dv}{dx} = \frac{1}{1+x^2}\)

Applying the product rule:

\(\frac{d}{dx}[(1+x^2) \tan^{-1}(x)] = (1+x^2)\frac{1}{1+x^2} + \tan^{-1}(x)(2x)\)

\(\frac{d}{dx}[(1+x^2) \tan^{-1}(x)] = 1 + 2x \tan^{-1}(x)\)

Now, differentiate x:

\(\frac{d}{dx}[x] = 1\)

Putting it all together:

\(\frac{dy}{dx} = [1 + 2x \tan^{-1}(x)] - 1\)

\(\frac{dy}{dx} = 2x \tan^{-1}(x)\)

Answer: \(2x \tan^{-1}(x)\)