Topic 2.10 Notes – Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

The Derivatives of tan x, cot x, sec x, and csc x

All trig derivatives on the AP exam assume x is in radians. If it were degrees, the formulas would look different.

Here are the four you must know cold:

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

$$\frac{d}{dx}(\cot x) = -\csc^2 x$$

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

$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

You already know:

• $\frac{d}{dx}(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) = -\sin x$

These new four can be derived from sine and cosine using identities and the quotient rule, but on a quiz or no‑calculator multiple choice, there isn’t time for that. You should recognize them instantly.

Two that students mix up every year:

• $(\tan x)' = \sec^2 x$
• $(\sec x)' = \sec x \tan x$

The square only goes with sec when it comes from tangent.

Rewriting Trig Functions Using Identities

Before differentiating, check if rewriting makes it easier.

Key identities:

• $\tan x = \dfrac{\sin x}{\cos x}$
• $\cot x = \dfrac{\cos x}{\sin x}$
• $\sec x = \dfrac{1}{\cos x}$
• $\csc x = \dfrac{1}{\sin x}$

Sometimes rewriting avoids messy quotient rule work.

For example:

$$f(x) = \frac{\cot x}{\csc x}$$

Rewrite first:

$$\frac{\cos x / \sin x}{1 / \sin x}$$

Multiply by the reciprocal:

$$= \cos x$$

Now the derivative is just $-\sin x$. Way easier.

This idea shows up a lot in free-response questions where they want to see if you simplify before differentiating. Recognizing structure is part of the skill here.

Applying the Derivative Rules

These functions rarely appear alone. They usually come wrapped inside something.

Sums and Constant Multiples

If

$$f(x) = 4\tan x - 7\sec x$$

Differentiate term by term:

$$f'(x) = 4\sec^2 x - 7\sec x \tan x$$

Nothing fancy. Just use the rules directly.

Chain Rule with Trig Functions

This is very common on tests.

If

$$f(x) = \sec(5x)$$

Think outer function first:

• Derivative of $\sec(\text{something})$ is $\sec(\text{something})\tan(\text{something})$

Then multiply by derivative of inside:

• Derivative of $5x$ is 5

So:

$$f'(x) = 5\sec(5x)\tan(5x)$$

General pattern to memorize:

• $\frac{d}{dx}[\tan(g(x))] = \sec^2(g(x)) \cdot g'(x)$
• $\frac{d}{dx}[\csc(g(x))] = -\csc(g(x))\cot(g(x)) \cdot g'(x)$

If your answer doesn’t include $g'(x)$, you forgot the chain rule.

Product Rule

These trig functions love to multiply each other.

If

$$f(x) = x \sec x$$

Use:

$$(fg)' = f'g + fg'$$

So:

$$f'(x) = 1\cdot \sec x + x(\sec x \tan x)$$

On a free-response question, don’t skip steps. Write the product rule structure clearly.

Quotient Rule

If

$$f(x) = \frac{\tan x}{x^2}$$

Use:

$$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$$

But pause first. Could you rewrite as $\tan x \cdot x^{-2}$ and use the product rule instead? Sometimes that’s cleaner.

Being flexible here saves algebra mistakes.

Powers of Trig Functions

If

$$f(x) = \cot^3 x$$

Rewrite as:

$$(\cot x)^3$$

Use chain rule:

$$f'(x) = 3(\cot x)^2(-\csc^2 x)$$

Don’t forget the negative from $(\cot x)'$. Missing that sign is a very common point loss.

Visual Reminder of Relationships

Seeing them together on the unit circle helps organize them in your head. In the diagram below, focus on how each trig function is represented by a segment tied to the angle $\theta$.

Unit circle definitions of the six trig functions

Notice:

• $\tan$ pairs naturally with $\sec$
• $\cot$ pairs naturally with $\csc$

That pairing shows up in their derivatives too.