AP Calculus AB and BC

Derivative of tan x: Answer, Proof, and Mistakes

Q: Is the derivative of \tan x \sec^2 x or \frac{1}{\cos^2 x}?

Both, because they are the same expression. Since \sec x = \frac{1}{\cos x}, squaring gives \sec^2 x = \frac{1}{\cos^2 x}. Use whichever form fits the rest of your work.

Q: Why is the derivative of \tan x always positive?

Wherever it is defined, \sec^2 x = \frac{1}{\cos^2 x} is 1 divided by a nonzero square, so it is always positive, never zero or negative. And since \cos^2 x \leq 1, dividing 1 by it gives \sec^2 x \geq 1. Either way the slope is positive, which is the calculus reason \tan x is increasing on every interval between its asymptotes.

Q: What is the second derivative of \tan x?

Differentiate \sec^2 x with the chain rule: \frac{d}{dx}[(\sec x)^2] = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x.

Q: Where is \tan x not differentiable?

At the odd multiples of \frac{\pi}{2}, that is x = \frac{\pi}{2} + n\pi for any integer n. There \cos x = 0, so \tan x has a vertical asymptote and is not even defined, let alone differentiable.

The derivative of tan x is sec^2 x, which equals 1/cos^2 x. It holds everywhere tan x is defined, meaning every x except odd multiples of pi/2 (where cos x = 0). You get it by applying the quotient rule to tan x = sin x / cos x, using the Pythagorean identity sin^2 x + cos^2 x = 1.

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

The proof, by the quotient rule

Rewrite the tangent as a quotient of sine and cosine, then differentiate with the quotient rule. Recall that $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$ .

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

The two negatives in $-\sin x \cdot (-\sin x)$ make a positive, so the numerator becomes $\cos^2 x + \sin^2 x$ , which equals $1$ by the Pythagorean identity.

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

Same answer, two forms

$\sec^2 x$ means $(\sec x)^2$ , and since $\sec x = \frac{1}{\cos x}$ , that is exactly $\frac{1}{\cos^2 x}$ . Both forms name the same result. It holds only where $\cos x \neq 0$ , so it excludes $x = \frac{\pi}{2} + n\pi$ , the odd multiples of $\frac{\pi}{2}$ where $\tan x$ is undefined.

Where it shows up on the AP exam

The College Board lists this derivative directly in Unit 2, Topic 2.10, Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions. Unit 2, Differentiation: Definition and Fundamental Properties, makes up 10 to 15% of the AB exam and 5 to 10% of BC.

The derivation leans on Topic 2.9, The Quotient Rule, and the ingredients come from Topic 2.7, Derivatives of cos x, sin x, e^x, and ln x. On the exam you rarely differentiate a bare $\tan x$ ; it usually sits inside a composite that needs the chain rule from Unit 3, Topic 3.1.

Chain-rule composites such as $\tan(3x)$ or $\tan(x^2)$ , where the answer is $\sec^2(u)$ times $u'$ for inside function $u$ .
Related rates and motion problems where an angle changes, so a $\tan$ term is differentiated with respect to time.
Tangent-line and slope questions that ask for $f'$ at a point when $f$ contains $\tan x$ .
Free-response work that expects you to recognize $\sec^2 x$ instantly, without pausing to re-derive it.

Common mistakes

Mistake	Why it is wrong
Writing $\frac{d}{dx}[\tan x] = \sec x \tan x$	That is the derivative of $\sec x$ , not $\tan x$ . The tangent gives $\sec^2 x$ .
$\frac{d}{dx}[\tan(3x)] = \sec^2(3x)$	The chain rule is missing. The inner derivative makes it $3\sec^2(3x)$ .
Reading $\sec^2 x$ as $\sec(x^2)$	$\sec^2 x$ means $(\sec x)^2$ , a squared output, not a squared input.
Numerator $\cos^2 x - \sin^2 x$	A sign slip. Since $\frac{d}{dx}[\cos x] = -\sin x$ , the term is $-\sin x \cdot (-\sin x) = +\sin^2 x$ , so the sum gives $1$ .
Using the formula at $x = \frac{\pi}{2}$	Both $\tan x$ and $\sec^2 x$ are undefined there. Check the domain before you evaluate.

Chain-rule composites, quick practice

Every composite follows one pattern: differentiate the outer tangent to $\sec^2(u)$ , then multiply by $u'$ , the derivative of the inside function $u$ .

\frac{d}{dx}\left[\tan(5x)\right] = \sec^2(5x) \cdot 5 = 5\sec^2(5x)

\frac{d}{dx}\left[\tan(x^2 + 1)\right] = \sec^2(x^2 + 1) \cdot 2x = 2x\sec^2(x^2 + 1)

For a product such as $x\tan x$ , pair this with the product rule: $\frac{d}{dx}[x\tan x] = \tan x + x\sec^2 x$ .

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

Is the derivative of $\tan x$ $\sec^2 x$ or $\frac{1}{\cos^2 x}$ ?

Both, because they are the same expression. Since $\sec x = \frac{1}{\cos x}$ , squaring gives $\sec^2 x = \frac{1}{\cos^2 x}$ . Use whichever form fits the rest of your work.

Why is the derivative of $\tan x$ always positive?

Wherever it is defined, $\sec^2 x = \frac{1}{\cos^2 x}$ is $1$ divided by a nonzero square, so it is always positive, never zero or negative. And since $\cos^2 x \leq 1$ , dividing $1$ by it gives $\sec^2 x \geq 1$ . Either way the slope is positive, which is the calculus reason $\tan x$ is increasing on every interval between its asymptotes.

What is the second derivative of $\tan x$ ?

Differentiate $\sec^2 x$ with the chain rule: $\frac{d}{dx}[(\sec x)^2] = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x$ .

Where is $\tan x$ not differentiable?

At the odd multiples of $\frac{\pi}{2}$ , that is $x = \frac{\pi}{2} + n\pi$ for any integer $n$ . There $\cos x = 0$ , so $\tan x$ has a vertical asymptote and is not even defined, let alone differentiable.