AP Calculus AB and BC

Derivative of x^2 arctan x: Product Rule

Q: What is the derivative of x^2 arctan x?

It is 2x\arctan x + \frac{x^{2}}{1+x^{2}}.

Q: Does the second term vanish at infinity?

No, it tends to 1. The first term is what dominates.

The derivative of x squared times arctan x is 2x arctan x plus x squared over 1 plus x squared. The product rule differentiates each factor in turn, and the second term uses the arctangent derivative.

\frac{d}{dx}\left[x^{2}\arctan x\right] = 2x\arctan x + \frac{x^{2}}{1+x^{2}}

The product rule

\frac{d}{dx}\left(x^{2}\arctan x\right) = 2x\arctan x + x^{2}\cdot\frac{1}{1+x^{2}}

Both terms survive: neither factor differentiates to zero, so nothing collapses.

Behaviour at infinity

The second term tends to $1$ while the first grows without bound, since arctangent flattens at $\frac{\pi}{2}$ and $2x$ does not. So the derivative behaves like $\pi x$ far out.

Common mistakes

Forgetting the second term.
Writing the arctangent derivative as $\frac{1}{1-x^{2}}$ , which belongs to a different function.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of x^2 arctan x?

It is $2x\arctan x + \frac{x^{2}}{1+x^{2}}$ .

Does the second term vanish at infinity?

No, it tends to $1$ . The first term is what dominates.