AP Calculus AB and BC

Derivative of sec(x^2): Chain Rule

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

It is 2x\sec\left(x^{2}\right)\tan\left(x^{2}\right).

Q: How is it different from sec^2 x?

One squares the input, the other the output. \sec^{2}x differentiates to 2\sec^{2}x\tan x.

The derivative of secant of x squared is 2x times secant of x squared times tangent of x squared. The secant derivative keeps its argument unchanged, and the chain rule multiplies by 2x, the derivative of the inside.

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

The chain rule

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

Both trigonometric factors keep the argument $x^{2}$ . Changing either to $x$ is the error that separates a right answer from a wrong one.

Not the same as sec squared x

$\sec\left(x^{2}\right)$ squares the INPUT; $\sec^{2}x$ squares the OUTPUT. The second is the derivative of tangent and differentiates to $2\sec^{2}x\tan x$ , with no factor of $x$ anywhere.

Common mistakes

Reading $\sec\left(x^{2}\right)$ as $\sec^{2}x$ .
Dropping the $2x$ chain rule factor.

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(x^2)?

It is $2x\sec\left(x^{2}\right)\tan\left(x^{2}\right)$ .

How is it different from sec^2 x?

One squares the input, the other the output. $\sec^{2}x$ differentiates to $2\sec^{2}x\tan x$ .