AP Calculus AB and BC

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

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

It is 2x\sec^2(x^2), equivalently \frac{2x}{\cos^2(x^2)}. The 2x is the inner derivative supplied by the chain rule.

Q: How is \tan(x^2) different from \tan^2 x?

\tan(x^2) is tangent of a squared input, with derivative 2x\sec^2(x^2). \tan^2 x is a squared tangent, with derivative 2\tan x\sec^2 x.

Q: Where is the derivative of \tan(x^2) undefined?

Wherever \cos(x^2) = 0, that is x = \pm\sqrt{\frac{\pi}{2}+k\pi} for whole numbers k. The function itself has vertical asymptotes at those same inputs.

The derivative of tan(x^2) is 2x sec^2(x^2). In prime notation, if f(x) = tan(x^2) then f'(x) = 2x sec^2(x^2). The chain rule turns the outer tangent into sec^2 of the inside, keeps x^2 inside, then multiplies by the inner derivative 2x.

\frac{d}{dx}\left[\tan(x^2)\right] = 2x\sec^2(x^2)

How to differentiate tan(x^2)

The outer function is tangent, the inner function is $x^2$ . Apply $\frac{d}{dx}\tan u = \sec^2 u\cdot u'$ with $u = x^2$ , so $u' = 2x$ .

\frac{d}{dx}\tan(x^2) = \sec^2(x^2)\cdot 2x = 2x\sec^2(x^2)

The inside never comes out of the secant. Only the inner derivative $2x$ moves to the front, and in cosine form the answer reads $\frac{2x}{\cos^2(x^2)}$ .

Why tan(x^2) is not the same as tan^2 x

$\tan(x^2)$ squares the input first and then takes the tangent. $\tan^2 x$ takes the tangent first and then squares it. Different compositions, different derivatives.

Function	Derivative
$\tan(x^2)$	$2x\sec^2(x^2)$
$\tan^2 x$	$2\tan x\sec^2 x$

The graphs differ too. $\tan(x^2)$ blows up wherever $x^2 = \frac{\pi}{2}+k\pi$ , so its vertical asymptotes crowd closer together as $|x|$ grows, while $\tan^2 x$ keeps evenly spaced asymptotes and stays nonnegative.

Common mistakes with the derivative of tan(x^2)

Answering $\sec^2(x^2)$ and dropping the inner derivative $2x$ .
Answering $2x\sec^2 x$ , differentiating the inside but forgetting to leave $x^2$ inside the secant.
Reading the function as $\tan^2 x$ and answering $2\tan x\sec^2 x$ .
Writing $\sec^2(2x)$ , which differentiates the inside in the wrong place entirely.

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 $\tan(x^2)$ ?

It is $2x\sec^2(x^2)$ , equivalently $\frac{2x}{\cos^2(x^2)}$ . The $2x$ is the inner derivative supplied by the chain rule.

How is $\tan(x^2)$ different from $\tan^2 x$ ?

$\tan(x^2)$ is tangent of a squared input, with derivative $2x\sec^2(x^2)$ . $\tan^2 x$ is a squared tangent, with derivative $2\tan x\sec^2 x$ .

Where is the derivative of $\tan(x^2)$ undefined?

Wherever $\cos(x^2) = 0$ , that is $x = \pm\sqrt{\frac{\pi}{2}+k\pi}$ for whole numbers $k$ . The function itself has vertical asymptotes at those same inputs.