Derivatives of tan, cot, sec, csc

AP Calculus AB· AP Calculus AB CED — Differentiation: Definition and Fundamental Properties· 14 min read

1. Deriving Formulas With the Quotient Rule★★☆☆☆⏱ 4 min

All four trigonometric functions studied here are defined as ratios of sine and cosine, so their derivatives can be derived using the quotient rule and the known derivatives of $ $an d$ oxed{ ext{cos}} $. R ec a l l t h e q u o t i e n t r u l e f or$ f(x) = \frac{g(x)}{h(x)}$:

f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}

To derive the derivative of $ $, s t a r tw i t h t h e i d e n t i t y$ = \frac{\sin x}{\cos x}$. Substituting into the quotient rule gives:

f'(x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}

Using the Pythagorean identity $ ^2 x + \sin^2 x = 1 $, t hi ss im pl i f i es t o$ \frac{1}{\cos^2 x} = \sec^2 x $, so$ \frac{d}{dx}[\tan x] = \sec^2 x$. Repeating this process for the other three functions gives the full set of standard formulas:

$\frac{d}{d x} [cot x] = - csc^{2} x$ (negative sign for co-functions)
$\frac{d}{d x} [sec x] = sec x tan x$
$\frac{d}{d x} [csc x] = - csc x cot x$ (negative sign for the co-function)

📐 Worked Example

Derive the derivative of $f (x) = cot x$ using the quotient rule and trigonometric identities, showing all steps.

1
Rewrite $cot x$ in terms of sine and cosine: By definition, $cot x = \frac{c o s x}{s i n x}$ , so $g (x) = cos x$ (numerator) and $h (x) = sin x$ (denominator).
2
Compute derivatives of the numerator and denominator:
3
$g'(x) = \frac{d}{dx}[\cos x] = -\sin x, \quad h'(x) = \frac{d}{dx}[\sin x] = \cos x$
4
Apply the quotient rule:
5
$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x}$
6
Simplify using the Pythagorean identity, factoring out the negative sign in the numerator:
7
$\frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$
8
Final result: $\frac{d}{d x} [cot x] = - csc^{2} x$ .

Exam tip:

If you blank out on a formula during the exam, you can re-derive any of these four derivatives in under one minute using the quotient rule, which is always acceptable for full credit.

2. Evaluating Derivatives at a Point★★☆☆☆⏱ 3 min

The most common routine AP exam question asks you to differentiate a linear combination of these trigonometric functions, then evaluate the derivative at a specific input to find the slope of a tangent line or instantaneous rate of change. This requires correct application of the constant multiple rule and sum/difference rule, plus simplification using known trigonometric values for common angles.

📐 Worked Example

Find the slope of the tangent line to $f (x) = 3 sec x - 4 tan x$ at $x = \frac{π}{4}$ .

1
Differentiate term-by-term using the derivative formulas:
2
$f'(x) = 3(\sec x \tan x) - 4(\sec^2 x)$
3
Recall trigonometric values at $x = \frac{π}{4}$ : $sec (\frac{π}{4}) = 2$ , $tan (\frac{π}{4}) = 1$ , so $sec^{2} (\frac{π}{4}) = 2$ .
4
Substitute $x = \frac{π}{4}$ into the derivative:
5
$f'\left(\frac{\pi}{4}\right) = 3(\sqrt{2} \cdot 1) - 4(2) = 3\sqrt{2} - 8 \approx -3.76$
6
The slope of the tangent line equals the derivative at $x = \frac{π}{4}$ , so the final slope is $32 - 8$ (or approximately -3.76).

Exam tip:

Explicitly write down the trigonometric value for each function before substituting, to avoid mixing up values for reciprocal trig functions.

3. Differentiating Composite Functions (Chain Rule)★★★☆☆⏱ 4 min

Most non-routine problems on the AP exam involve composite functions, where one of the four trigonometric functions is the outer function of a more complex expression. Recall the chain rule: for $f (x) = u (v (x))$ , $f^{'} (x) = u^{'} (v (x)) \cdot v^{'} (x)$ . For a trigonometric outer function, you compute the derivative of the trig function, evaluate it at the inner function, then multiply by the derivative of the inner function.

📐 Worked Example

Find the derivative of $f (x) = csc (5 x^{2} + π x)$ .

1
Identify outer and inner functions for the chain rule: Let $u (v) = csc v$ (outer) and $v (x) = 5 x^{2} + π x$ (inner).
2
Differentiate the outer function using the cosecant derivative formula: $u^{'} (v) = - csc v cot v$ .
3
Differentiate the inner function using the power rule: $v^{'} (x) = 10 x + π$ .
4
Apply the chain rule, substituting back $v (x)$ for $v$ :
5
$f'(x) = -\csc(5x^2 + \pi x) \cot(5x^2 + \pi x) \cdot (10x + \pi)$

Exam tip:

Even if the question does not require you to simplify your final answer on an FRQ, always write the chain rule factor explicitly to earn full credit; omitting it will cost you a point even if the rest of the derivative is correct.

4. AP-Style Practice Problems★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding with this multiple-choice question:

If $y = tan (2 x) - csc (\frac{x}{2})$ , what is the value of $\frac{d y}{d x}$ at $x = π$ ?
- $- 2$
- $0$
- $2$
- $4$
Reveal answer
2 —
Correct. Differentiating term-by-term with the chain rule gives $\frac{d y}{d x} = 2 sec^{2} (2 x) + \frac{1}{2} csc (\frac{x}{2}) cot (\frac{x}{2})$ , which evaluates to 2 at $x = π$ .

📐 Worked Example

Let $f (x) = x sec x$ for $- \frac{π}{2} < x < \frac{π}{2}$ . (a) Find $f^{'} (x)$ . (b) Find the equation of the tangent line at $x = \frac{π}{4}$ . (c) Approximate $f (\frac{π}{4} + 0.02)$ using the tangent line.

1
Part (a): Use the product rule for $f (x) = x \cdot sec x$ :
2
$f'(x) = 1 \cdot \sec x + x \cdot \sec x \tan x = \sec x (1 + x \tan x)$
3
Part (b): Calculate $f (\frac{π}{4})$ and $f^{'} (\frac{π}{4}) :$
4
$f\left(\frac{\pi}{4}\right) = \frac{\pi \sqrt{2}}{4}, \quad f'\left(\frac{\pi}{4}\right) = \sqrt{2} + \frac{\pi \sqrt{2}}{4}$
5
Write the tangent line in point-slope form, then simplify:
6
$y - \frac{\pi\sqrt{2}}{4} = \left(\sqrt{2} + \frac{\pi\sqrt{2}}{4}\right)\left(x - \frac{\pi}{4}\right)$
7
$y = \sqrt{2}\left(1 + \frac{\pi}{4}\right)x - \frac{\pi\sqrt{2}}{4}$
8
Part (c): Substitute $x = \frac{π}{4} + 0.02$ to get the approximation:
9
$y \approx 1.16$

📐 Worked Example

The position of an oscillating block is $s (t) = 10 tan (0.1 t)$ , where $s$ is in cm and $t$ in seconds. Find the instantaneous velocity at $t = π$ and interpret the result.

1
Instantaneous velocity is the derivative of position. Apply the chain rule:
2
$v(t) = \frac{d}{dt}[10 \tan(0.1t)] = 10 \cdot \sec^2(0.1t) \cdot 0.1 = \sec^2(0.1t)$
3
Evaluate at $t = π$ , where $0.1 t = \frac{π}{10} :$
4
$v(\pi) = \sec^2\left(\frac{\pi}{10}\right) = \left(\frac{1}{\cos\left(\frac{\pi}{10}\right)}\right)^2 \approx 1.11$
5
Interpretation: At $t = π$ seconds, the block is moving up the track at ~1.11 cm per second.

5. Common Pitfalls

Wrong move:

Writing $\frac{d}{d x} [cot x] = csc^{2} x$ or $\frac{d}{d x} [csc x] = csc x cot x$ , dropping the negative sign on co-function derivatives.

Why:

Students forget that all co-trig derivatives have a negative sign from the quotient rule derivation, and mix up the sign pattern.

Correct move:

Memorize the pattern: all "co-" functions have negative derivatives; if you are unsure, quickly re-derive the formula to confirm the sign.

Wrong move:

When differentiating $tan (4 x)$ , writing $\frac{d y}{d x} = sec^{2} (4 x)$ , omitting the chain rule factor of 4.

Why:

Students memorize the derivative of $tan x$ and forget to multiply by the derivative of the inner linear term.

Correct move:

For any composite function $tan (u (x))$ , always write the derivative as $sec^{2} (u (x)) \cdot u^{'} (x)$ , explicitly adding the $u^{'} (x)$ term before moving on.

Wrong move:

Confusing derivative formulas, writing $\frac{d}{d x} [sec x] = sec^{2} x$ and $\frac{d}{d x} [tan x] = sec x tan x$ .

Why:

Similar notation leads to mixing up which formula pairs with which function.

Correct move:

If you mix up formulas, spend 30 seconds re-deriving the formula from the quotient rule instead of guessing.

Wrong move:

When differentiating $f (x) = x^{2} sec x$ , writing $f^{'} (x) = 2 x \cdot sec x tan x$ , forgetting to apply the product rule.

Why:

Students focus on the trigonometric derivative and ignore that it is multiplied by another function.

Correct move:

Always check for products, quotients, or composition before differentiating; use the product rule for any product of two functions, regardless of type.

Wrong move:

When simplifying the derivative of $tan x$ , writing $\frac{1}{c o s ^{2} x} = \frac{1}{s e c ^{2} x}$ instead of $sec^{2} x$ .

Why:

Students mix up the reciprocal relationship between cosine and secant, flipping the fraction incorrectly.

Correct move:

Remember that $\frac{1}{c o s ^{2} x} = (\frac{1}{c o s x})^{2} = sec^{2} x$ , so the reciprocal of $cos^{2} x$ is $sec^{2} x$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Derivative of tangent	$\frac{d}{d x} [tan x] = sec^{2} x$	Valid for all $x \neq = \frac{π}{2} + k π$ (integer $k$ )
Derivative of cotangent	$\frac{d}{d x} [cot x] = - csc^{2} x$	Valid for all $x \neq = k π$ (integer $k$ ); negative for co-function
Derivative of secant	$\frac{d}{d x} [sec x] = sec x tan x$	Valid for all $x \neq = \frac{π}{2} + k π$ (integer $k$ )
Derivative of cosecant	$\frac{d}{d x} [csc x] = - csc x cot x$	Valid for all $x \neq = k π$ (integer $k$ ); negative for co-function
Composite tangent	$\frac{d}{d x} [tan (u (x))] = sec^{2} (u (x)) \cdot u^{'} (x)$	Always multiply by chain rule factor $u^{'} (x)$
Composite cotangent	$\frac{d}{d x} [cot (u (x))] = - csc^{2} (u (x)) \cdot u^{'} (x)$	Do not forget negative sign or chain rule
Composite secant	$\frac{d}{d x} [sec (u (x))] = sec (u (x)) tan (u (x)) \cdot u^{'} (x)$	Avoid mixing up with cosecant formula
Composite cosecant	$\frac{d}{d x} [csc (u (x))] = - csc (u (x)) cot (u (x)) \cdot u^{'} (x)$	Negative sign for all co-function derivatives

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Evaluate derivative at a point
2022 · FRQ
Tangent line problem

What's Next

This topic is a critical building block for all subsequent work with trigonometric functions in AP Calculus AB. Mastery of these four derivative formulas is required for nearly every problem involving trigonometric functions in later units, including implicit differentiation, related rates, optimization, integration, and differential equations. Immediately next, you will apply these formulas when working with the full chain rule for all composite functions, regularly differentiating complex mixed trigonometric-algebraic functions. This topic also lays the foundation for integrating trigonometric functions later in the course, as integral formulas are just the reverse of the derivative formulas you learned here. Tangent line and instantaneous rate of change problems, heavily tested on the AP exam, rely entirely on this skill.