AP Calculus AB and BC

Derivative of sec^2 x: Answer, Proof, and Mistakes

Q: What is the derivative of \sec^{2}x?

It is 2\sec^{2}x\tan x, equivalently \frac{2\sin x}{\cos^{3}x}.

Q: Is the derivative of \tan^{2}x the same?

Yes, it is also 2\tan x\sec^{2}x. The identity \sec^{2}x = 1+\tan^{2}x means the two functions differ by the constant 1, and constants differentiate away.

Q: What is \int \sec^{2}x\,dx?

That is the reverse question, and the answer is \tan x + C, since \frac{d}{dx}\tan x = \sec^{2}x. Differentiating \sec^{2}x instead gives 2\sec^{2}x\tan x.

The derivative of sec^2 x is 2 sec^2(x) tan(x). In prime notation, if f(x) = sec^2 x then f'(x) = 2 sec^2(x) tan(x). Read sec^2 x as (sec x)^2 and use the chain rule: 2 sec x times the derivative of sec x, which is sec x tan x.

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

How to differentiate sec^2 x

The notation $\sec^2 x$ means $(\sec x)^2$ , so the outer function is the square and the inner function is $\sec x$ . The chain rule needs the standard fact $\frac{d}{dx}\sec x = \sec x\tan x$ .

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

In cosine form the same answer reads $\frac{2\sin x}{\cos^{3}x}$ , which is the version a calculator check will match.

Two checks that confirm the answer

First check: $\sec^{2}x$ is itself the derivative of $\tan x$ , so this result is the second derivative of tangent.

\frac{d}{dx}\tan x = \sec^{2}x, \qquad \frac{d^{2}}{dx^{2}}\tan x = 2\sec^{2}x\tan x

Second check: the Pythagorean identity $\sec^{2}x = 1+\tan^{2}x$ means $\sec^{2}x$ and $\tan^{2}x$ differ by the constant $1$ , so the two have the same derivative. Differentiating $\tan^{2}x$ by the chain rule gives $2\tan x\sec^{2}x$ , the same expression.

Sign follows $\tan x$ , so $\sec^{2}x$ decreases on $\left(-\frac{\pi}{2},0\right)$ , bottoms out at $1$ when $x=0$ , and increases on $\left(0,\frac{\pi}{2}\right)$ .

Common mistakes with the derivative of sec^2 x

Answering $2\sec x$ , applying the power rule to the outside and skipping the inner derivative $\sec x\tan x$ .
Dropping the $2$ and writing $\sec^{2}x\tan x$ . The square supplies that factor.
Confusing $\sec^{2}x$ with $\sec(x^{2})$ , whose derivative is $2x\sec(x^{2})\tan(x^{2})$ .
Reporting $\sec^{2}x$ again, which is the derivative of $\tan x$ , not of $\sec^{2}x$ .

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 $\sec^{2}x$ ?

It is $2\sec^{2}x\tan x$ , equivalently $\frac{2\sin x}{\cos^{3}x}$ .

Is the derivative of $\tan^{2}x$ the same?

Yes, it is also $2\tan x\sec^{2}x$ . The identity $\sec^{2}x = 1+\tan^{2}x$ means the two functions differ by the constant $1$ , and constants differentiate away.

What is $\int \sec^{2}x\,dx$ ?

That is the reverse question, and the answer is $\tan x + C$ , since $\frac{d}{dx}\tan x = \sec^{2}x$ . Differentiating $\sec^{2}x$ instead gives $2\sec^{2}x\tan x$ .