AP Calculus AB and BC

Derivative of sec x: Answer, Proof, and Mistakes

Q: Is the derivative of sec x equal to sec^2 x?

No. \sec^2 x is the derivative of \tan x. The derivative of \sec x is \sec x \tan x, the product of secant and tangent. The two are easy to swap, so lock in the pair \tan x \to \sec^2 x and \sec x \to \sec x \tan x.

Q: Why does the derivative of sec x have no minus sign, but csc x does?

In the proof, the power rule and the derivative of \cos x each contribute a minus sign, and the two cancel, so no minus sign survives. For \csc x = (\sin x)^{-1} the inner derivative is +\cos x, so nothing cancels and a leading minus sign remains: -\csc x \cot x. This is the co-function sign pattern.

Q: How do you differentiate sec(2x) or other composites?

Use the chain rule. Differentiate the outside as \sec(\text{inside})\tan(\text{inside}), then multiply by the derivative of the inside. For \sec(2x) the inside derivative is 2, giving 2\sec(2x)\tan(2x).

Q: Where is the derivative of sec x undefined?

At x = \frac{\pi}{2} + n\pi (the odd multiples of \frac{\pi}{2}), because \cos x = 0 there. At those points \sec x itself is undefined, so neither the function nor its derivative \sec x \tan x exists.

The derivative of sec x is sec x tan x. Write sec x as (cos x)^-1 and apply the chain rule: the outer power gives -(cos x)^-2 and the inner derivative of cos x is -sin x, so the two negatives cancel and leave sin x over cos^2 x, which regroups as sec x tan x.

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

Proving it with the chain rule

Start from the definition of secant. Since $\sec x = \frac{1}{\cos x} = (\cos x)^{-1}$ , the derivative is just a power of a function, so the chain rule (Topic 3.1) applies directly. The quotient rule (Topic 2.9) on $\frac{1}{\cos x}$ lands on the same result if you prefer that route.

\frac{d}{dx}\left[(\cos x)^{-1}\right] = -1\cdot(\cos x)^{-2}\cdot(-\sin x) = \frac{\sin x}{\cos^2 x}

The outer power rule contributes one negative sign and the derivative of the inner $\cos x$ contributes another, so the two cancel. Now regroup the single fraction into a product:

\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x

That cancellation is exactly why the derivative of $\sec x$ carries no leading minus sign, while its co-function $\csc x$ does.

The co-function sign pattern

Notice there is no minus sign in $\sec x \tan x$ . Set it beside its co-function: $\csc x$ differentiates to $-\csc x \cot x$ . Across all six trig derivatives, the three co-functions (cosine, cotangent, cosecant) are exactly the ones whose derivatives carry a leading negative sign.

Function $f(x)$	Derivative $f'(x)$	Leading sign
$\sec x$	$\sec x \tan x$	$+$
$\csc x$	$-\csc x \cot x$	$-$

So the sec and csc rules are mirror images. Keep $\sec x \tan x$ in memory, then swap each factor for its co-function and attach a minus sign to recover $\csc x \to -\csc x \cot x$ . The full six-row table lives in the trig derivatives guide.

Where sec x shows up on the AP exam

The secant derivative is introduced in Unit 2 (Differentiation: Definition and Fundamental Properties), specifically Topic 2.10, Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions. Unit 2 carries 10-15% of the AB exam and 5-10% of BC.

On the exam $\sec x$ rarely stands alone. It usually appears inside a composite such as $\sec(2x)$ or $\sec(x^2)$ , which pairs Topic 2.10 with the chain rule (Topic 3.1, Unit 3). It also turns up in tangent-line approximation (Unit 4, Topic 4.6) and related-rates work (Unit 4), where $\sec$ and $\tan$ show up together.

Common mistakes to avoid

Answering $\sec^2 x$ . That is the derivative of $\tan x$ , not of $\sec x$ . Keep the pair straight: $\tan x \to \sec^2 x$ and $\sec x \to \sec x \tan x$ .
Attaching a negative sign. The derivative of $\sec x$ has no leading minus sign; the minus belongs to its co-function, $\csc x \to -\csc x \cot x$ .
Dropping the inner derivative on composites. $\frac{d}{dx}\sec(3x) = 3\sec(3x)\tan(3x)$ , not $\sec(3x)\tan(3x)$ . The chain rule factor is required.
Writing only one factor. The answer is the product $\sec x \cdot \tan x$ ; leaving off either $\sec x$ or $\tan x$ loses half of it.
Treating $\sec x$ as $\frac{1}{\sec x}$ or as an inverse function. Recall $\sec x = \frac{1}{\cos x}$ , so its derivative comes from differentiating $(\cos x)^{-1}$ , not from a reciprocal or an arcsecant rule.

Two worked chain-rule composites

Every composite runs off one template: $\frac{d}{dx}\sec u = \sec u \tan u \cdot u'$ , where $u'$ is the derivative of whatever sits inside.

Linear inside. Let $u = 3x$ , so $u' = 3$ .

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

Power inside. Let $u = x^2$ , so $u' = 2x$ .

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

Is the derivative of sec x equal to sec^2 x?

No. $\sec^2 x$ is the derivative of $\tan x$ . The derivative of $\sec x$ is $\sec x \tan x$ , the product of secant and tangent. The two are easy to swap, so lock in the pair $\tan x \to \sec^2 x$ and $\sec x \to \sec x \tan x$ .

Why does the derivative of sec x have no minus sign, but csc x does?

In the proof, the power rule and the derivative of $\cos x$ each contribute a minus sign, and the two cancel, so no minus sign survives. For $\csc x = (\sin x)^{-1}$ the inner derivative is $+\cos x$ , so nothing cancels and a leading minus sign remains: $-\csc x \cot x$ . This is the co-function sign pattern.

How do you differentiate sec(2x) or other composites?

Use the chain rule. Differentiate the outside as $\sec(\text{inside})\tan(\text{inside})$ , then multiply by the derivative of the inside. For $\sec(2x)$ the inside derivative is $2$ , giving $2\sec(2x)\tan(2x)$ .

Where is the derivative of sec x undefined?

At $x = \frac{\pi}{2} + n\pi$ (the odd multiples of $\frac{\pi}{2}$ ), because $\cos x = 0$ there. At those points $\sec x$ itself is undefined, so neither the function nor its derivative $\sec x \tan x$ exists.