AP Calculus AB and BC

Derivative of arctan(x^2): Answer, Proof, Mistakes

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

It is \frac{2x}{1 + x^4}, from the chain rule with inner function x^2.

Q: Why is the denominator 1 + x^4 instead of 1 + x^2?

The rule is \frac{1}{1 + u^2} with u the inner function. Since u = x^2, the square is (x^2)^2 = x^4.

Q: Is the derivative of arctan(x^2) ever negative?

Yes, for every x < 0. The numerator 2x is negative there while 1 + x^4 is positive for all x.

The derivative of arctan(x^2) with respect to x is 2x/(1 + x^4). The chain rule gives 1 over 1 plus (x^2)^2, times the inner derivative 2x, and (x^2)^2 equals x^4, so the denominator is 1 + x^4 rather than 1 + x^2. The formula holds for every real number x.

\frac{d}{dx}\left[\arctan(x^2)\right] = \frac{2x}{1+x^4}

The proof: inverse trig rule plus the chain rule

Topic 3.4 supplies $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ , and the chain rule attaches an inner derivative whenever the input is something other than a bare $x$ .

\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}

Here $u = x^2$ , so $u' = 2x$ and $u^2 = (x^2)^2 = x^4$ .

\frac{d}{dx}\arctan(x^2) = \frac{2x}{1 + (x^2)^2} = \frac{2x}{1 + x^4}

Squaring the inner function doubles its exponent, which is the whole reason an $x^4$ shows up in a problem that started with $x^2$ .

What the derivative says about the graph

$\arctan(x^2)$ is an even function, so its derivative is odd: negative for $x < 0$ , zero at $x = 0$ , positive for $x > 0$ . The one critical point sits at the origin and is the absolute minimum, with value $\arctan 0 = 0$ .

As $|x|$ grows, $\arctan(x^2)$ approaches the horizontal asymptote $y = \frac{\pi}{2}$ , and the derivative $\frac{2x}{1 + x^4}$ shrinks roughly like $\frac{2}{x^3}$ , so the curve flattens fast.

Common mistakes

Answering $\frac{1}{1 + x^4}$ and leaving out the inner derivative $2x$ .
Answering $\frac{2x}{(1 + x^2)^2}$ , squaring the whole denominator instead of squaring the inner function inside it.
Answering $\frac{2x}{1 + x^2}$ , which uses the plain $\arctan x$ rule and forgets that $u^2$ means $(x^2)^2$ .
Confusing $\arctan(x^2)$ with $(\arctan x)^2$ , whose derivative is $\frac{2\arctan x}{1 + x^2}$ .

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 arctan(x^2)?

It is $\frac{2x}{1 + x^4}$ , from the chain rule with inner function $x^2$ .

Why is the denominator 1 + x^4 instead of 1 + x^2?

The rule is $\frac{1}{1 + u^2}$ with $u$ the inner function. Since $u = x^2$ , the square is $(x^2)^2 = x^4$ .

Is the derivative of arctan(x^2) ever negative?

Yes, for every $x < 0$ . The numerator $2x$ is negative there while $1 + x^4$ is positive for all $x$ .