Derivatives of
Reciprocal Trig Functions
Master the derivatives of \(\tan x\), \(\cot x\), \(\sec x\), and \(\csc x\) using quotient rule connections and pattern recognition.
AP Calculus AB
Why These Four Derivatives Matter
You already know \(\frac{d}{dx}[\sin x] = \cos x\) and \(\frac{d}{dx}[\cos x] = -\sin x\). But calculus problems rarely stop at sine and cosine. You need the derivatives of all six trig functions to handle real-world modeling, integration, and chain-rule problems on the AP exam.
The Building Blocks: Reciprocal Definitions
\(\tan x\)
$$\frac{\sin x}{\cos x}$$
\(\cot x\)
$$\frac{\cos x}{\sin x}$$
\(\sec x\)
$$\frac{1}{\cos x}$$
\(\csc x\)
$$\frac{1}{\sin x}$$
Visualizing \(\tan x\) and Its Derivative
Deriving \(\frac{d}{dx}[\tan x]\)
1
Rewrite as a quotient: \(\tan x = \dfrac{\sin x}{\cos x}\)
2
Apply quotient rule: \(\dfrac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}\)
3
Simplify numerator: \(\dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x}\)
4
Final result: \(\dfrac{d}{dx}[\tan x] = \sec^2 x\)
A light-and-camera studio with exact sightlines
A rotating light turns angle change into motion along a wall. Differentiate the right-triangle relationship to connect the two rates.Spotlight: angle from the wall normal
A rotating light turns angle change into motion along a wall. Differentiate the right-triangle relationship to connect the two rates.The Complete Set of Four Formulas
\(\frac{d}{dx}[\tan x]\)
$$\sec^2 x$$
\(\frac{d}{dx}[\cot x]\)
$$-\csc^2 x$$
\(\frac{d}{dx}[\sec x]\)
$$\sec x \tan x$$
\(\frac{d}{dx}[\csc x]\)
$$-\csc x \cot x$$
\(\tan x\) vs \(\cot x\): The Negative Sign
| \(\tan x\) | \(\cot x\) | |
|---|---|---|
| Definition | \(\dfrac{\sin x}{\cos x}\) | \(\dfrac{\cos x}{\sin x}\) |
| Derivative | \(\sec^2 x\) | \(-\csc^2 x\) |
| Sign | Positive | Negative |
| Always increasing? | Yes (on each branch) | Always decreasing |
\(\sec x\) vs \(\csc x\): Same Structure, Different Sign
| \(\sec x\) | \(\csc x\) | |
|---|---|---|
| Definition | \(\dfrac{1}{\cos x}\) | \(\dfrac{1}{\sin x}\) |
| Derivative | \(\sec x \tan x\) | \(-\csc x \cot x\) |
| Structure | (itself) × (partner) | -(itself) × (partner) |
How to Apply These Derivatives
1
Identify the outer trig function: Is it \(\tan\), \(\cot\), \(\sec\), or \(\csc\)?
2
Recall the base derivative: Use the four formulas from the previous slide.
3
Check for a chain rule: If the argument is not just \(x\), multiply by the derivative of the inner function.
4
Simplify: Look for Pythagorean identity simplifications or common factors.
Example 1: Basic Derivative
Find \(\dfrac{d}{dx}[3\tan x - 2\sec x]\).
1
Differentiate term by term: \(\frac{d}{dx}[3\tan x] - \frac{d}{dx}[2\sec x]\)
2
Apply formulas: \(3\sec^2 x - 2\sec x \tan x\)
3
Factor (optional): \(\sec x(3\sec x - 2\tan x)\)
Example 2: Chain Rule Application
Find \(\dfrac{d}{dx}[\csc(5x^2)]\).
1
Identify: outer = \(\csc(u)\), inner = \(u = 5x^2\)
2
Base derivative: \(\frac{d}{du}[\csc u] = -\csc u \cot u\)
3
Inner derivative: \(u' = 10x\)
4
Multiply: \(-\csc(5x^2)\cot(5x^2) \cdot 10x\)
Example 3: Product Rule Meets Trig
Find \(\dfrac{d}{dx}[x^2 \sec x]\).
1
Product rule: \(f = x^2\), \(g = \sec x\)
2
Compute: \(f' = 2x\), \(g' = \sec x \tan x\)
3
Apply \(f'g + fg'\): \(2x\sec x + x^2 \sec x \tan x\)
4
Factor: \(x\sec x(2 + x\tan x)\)
Proof: Deriving \(\frac{d}{dx}[\sec x]\)
1
Rewrite: \(\sec x = \dfrac{1}{\cos x} = (\cos x)^{-1}\)
2
Quotient rule: \(\dfrac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x}\)
3
Split the fraction: \(\dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x}\)
4
Recognize: \(\sec x \cdot \tan x\)
The Negative Sign Pattern
All six trig derivatives follow a single rule about where the negative sign lands:
Positive derivatives
\(\frac{d}{dx}[\sin x] = \cos x\)
\(\frac{d}{dx}[\tan x] = \sec^2 x\)
\(\frac{d}{dx}[\sec x] = \sec x \tan x\)
\(\frac{d}{dx}[\tan x] = \sec^2 x\)
\(\frac{d}{dx}[\sec x] = \sec x \tan x\)
Negative derivatives
\(\frac{d}{dx}[\cos x] = -\sin x\)
\(\frac{d}{dx}[\cot x] = -\csc^2 x\)
\(\frac{d}{dx}[\csc x] = -\csc x \cot x\)
\(\frac{d}{dx}[\cot x] = -\csc^2 x\)
\(\frac{d}{dx}[\csc x] = -\csc x \cot x\)
Common Mistakes to Avoid
✕ Forgetting the Negative
Writing \(\frac{d}{dx}[\cot x] = \csc^2 x\) instead of \(-\csc^2 x\). The "co-" functions always carry a negative.
✕ Dropping the Chain Rule
Writing \(\frac{d}{dx}[\sec(3x)] = \sec(3x)\tan(3x)\) without the extra factor of \(3\).
✕ Mixing Up Squared Forms
Confusing \(\sec^2 x\) (derivative of \(\tan x\)) with \(\csc^2 x\) (derivative of \(\cot x\)). Remember: \(\tan\) pairs with \(\sec\), \(\cot\) pairs with \(\csc\).
Summary: All Six Trig Derivatives
| Function | Derivative | Sign |
|---|---|---|
| \(\sin x\) | \(\cos x\) | + |
| \(\cos x\) | \(-\sin x\) | \(-\) |
| \(\tan x\) | \(\sec^2 x\) | + |
| \(\cot x\) | \(-\csc^2 x\) | \(-\) |
| \(\sec x\) | \(\sec x \tan x\) | + |
| \(\csc x\) | \(-\csc x \cot x\) | \(-\) |
Question 1 of 5
What is \(\dfrac{d}{dx}[\cot x]\)?
The derivative of \(\cot x\) is \(-\csc^2 x\). Since \(\cot x\) is a "co-" function, its derivative carries a negative sign. Also, \(\cot\) pairs with \(\csc\), not \(\sec\).
Question 2 of 5
Find \(\dfrac{d}{dx}[\sec(4x)]\).
By the chain rule: \(\frac{d}{dx}[\sec(4x)] = \sec(4x)\tan(4x) \cdot 4 = 4\sec(4x)\tan(4x)\). Choice (a) is missing the chain rule factor. Choice (d) confuses the derivative of \(\sec\) with the derivative of \(\tan\).
Question 3 of 5
Which of the following is \(\dfrac{d}{dx}[x \cdot \csc x]\)?
Using the product rule with \(f = x\) and \(g = \csc x\): \(f'g + fg' = (1)(\csc x) + (x)(-\csc x \cot x) = \csc x - x\csc x \cot x\). Choice (b) forgets the \(f'g\) term. Choice (c) has the wrong sign on the second term.
Question 4 of 5
If \(f(x) = \tan^2 x\), what is \(f'(x)\)?
Using the chain rule with the power rule: \(\frac{d}{dx}[\tan^2 x] = 2\tan x \cdot \frac{d}{dx}[\tan x] = 2\tan x \cdot \sec^2 x\). Choice (a) forgets the derivative of the inner function. Choice (c) forgets the \(\tan x\) factor from the power rule.
Question 5 of 5
Which derivative is always negative wherever it is defined?
Since \(\csc^2 x\) is always positive (a squared value), \(-\csc^2 x\) is always negative. This means \(\cot x\) is always decreasing on every interval of its domain. Choice (a) is always positive. Choices (c) and (d) can be positive or negative depending on the quadrant.