AP Calculus AB and BC

Derivative of tan 3x: Answer and Chain Rule

Q: Is it always positive?

Yes, wherever it exists, because \sec^{2} is a square. So \tan 3x increases on each interval between asymptotes.

The derivative of tan 3x is 3 times sec squared of 3x. Tangent differentiates to secant squared with the argument unchanged, and the chain rule multiplies by 3 for the inner function.

\frac{d}{dx}\left[\tan 3x\right] = 3\sec^{2} 3x

Applying the chain rule

\frac{d}{dx}\tan(3x) = \sec^{2}(3x)\cdot 3 = 3\sec^{2}3x

The derivative is always positive where it exists, since a square cannot be negative, so $\tan 3x$ is increasing on every interval between its asymptotes.

Asymptotes come three times as often

$\tan 3x$ is undefined when $3x$ is an odd multiple of $\frac{\pi}{2}$ , so the asymptotes sit at $x = \frac{\pi}{6} + \frac{k\pi}{3}$ , three times as densely as for $\tan x$ .

Common mistakes

Answering $\sec^{2}3x$ with no chain rule factor.
Writing $3\sec^{2}x$ , differentiating the inner function into the argument.
Reading $\sec^{2}3x$ as $\sec\left(9x^{2}\right)$ . The square applies to the function value.

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 3x?

It is $3\sec^{2}3x$ .

Is it always positive?

Yes, wherever it exists, because $\sec^{2}$ is a square. So $\tan 3x$ increases on each interval between asymptotes.