Difficulty: Intermediate | Prerequisites: Product rule, quotient rule, basic trig identities, unit circle.
This material combines the differentiation rules you have already learnt (product rule, quotient rule) with the six trigonometric functions. The result is a toolbox that lets you differentiate any expression built from polynomials and trig functions. You will use these derivatives constantly from here through integration, so they need to become automatic. This section also asks you to prove one derivative formula from scratch, which tests both your algebra and your understanding of how the rules connect.
You need to know the derivatives of sin, cos, tan, sec, csc and cot by heart, and you need to apply the product and quotient rules cleanly when these functions appear in combinations. Most exam errors come from sign mistakes and from misapplying the quotient rule, not from the trig itself.
Product rule
(f · g)' = f' · g + f · g'. In simple terms, differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Quotient rule
(f / g)' = (f' · g − f · g') / g². Think of it as "low d-high minus high d-low, over the square of what's below."
Derivative of sin(x)
(sin x)' = cos x.
Derivative of cos(x)
(cos x)' = −sin x. Note the negative sign.
Derivative of tan(x)
(tan x)' = sec²(x). Derived from the quotient rule applied to sin(x)/cos(x).
Derivative of sec(x)
(sec x)' = sec(x) tan(x).
Tangent line
The line through the point (a, f(a)) with slope f'(a). Equation: y − f(a) = f'(a)(x − a). In simple terms, the tangent line is the best straight-line approximation to the curve at that point.
(sin x)' = cos x
(cos x)' = −sin x
(tan x)' = sec²x
(sec x)' = sec x tan x
(csc x)' = −csc x cot x
(cot x)' = −csc²x
Notice the pattern: the "co-" functions (cos, csc, cot) all pick up a negative sign in their derivatives.
Example (a): f(x) = ³√x · sec(x)
Use the product rule with f(x) = x^(1/3) and g(x) = sec(x).
f'(x) = (1/3)x^(−2/3) · sec(x) + x^(1/3) · sec(x)tan(x)
Simplified: (1/3)x^(−2/3) sec(x) + x^(1/3) sec(x) tan(x).
Example (b): g(x) = x² cos(x) − x^(1/4)
Differentiate term by term. For x² cos(x), use the product rule:
g'(x) = [2x cos(x) + x²(−sin x)] − (1/4)x^(−3/4)
= 2x cos(x) − x² sin(x) − (1/4)x^(−3/4).
Example (c): h(x) = x³ / (10 + tan x)
Quotient rule with f(x) = x³, g(x) = 10 + tan(x):
h'(x) = [3x²(10 + tan x) − x³ sec²x] / (10 + tan x)²
= [3x²(10 + tan x) − x³ sec²x] / (10 + tan x)².
Example (d): p(t) = (2 tan t + 2) / sec(t)
Quotient rule:
p'(t) = [2 sec²(t) · sec(t) − (2 tan t + 2) · sec(t)tan(t)] / sec²(t).
Factor sec(t) from the numerator, cancel with the denominator:
= 2(1 + tan t) / sec(t).
Alternatively, simplify first: (2 tan t + 2)/sec(t) = 2(tan t + 1) cos(t) = 2 sin(t) + 2 cos(t), then differentiate directly to get 2 cos(t) − 2 sin(t).
Example (e): f(x) = −3x sin(x) cos(x)
This is a product of three functions. Treat it as (−3x) · (sin x · cos x) or apply the product rule twice.
f'(x) = −3[sin(x)cos(x) + x cos²(x) − x sin²(x)]
= −3 sin(x)cos(x) − 3x cos(2x).
The last step uses the identity cos²x − sin²x = cos(2x).
Write tan(x) = sin(x)/cos(x) and apply the quotient rule:
(tan x)' = [cos(x) · cos(x) − sin(x) · (−sin(x))] / cos²(x)
= [cos²(x) + sin²(x)] / cos²(x)
= 1 / cos²(x)
= sec²(x).
The Pythagorean identity cos²x + sin²x = 1 does the heavy lifting in the numerator. This derivation is a common "show your work" exam question.
Find the equation of the tangent line to f(x) = cos(x) + sec(x) at x = π.
Step 1, find f(π): cos(π) + sec(π) = −1 + (−1) = −2. The point is (π, −2).
Step 2, find f'(x): f'(x) = −sin(x) + sec(x)tan(x).
Step 3, evaluate f'(π): −sin(π) + sec(π)tan(π) = 0 + (−1)(0) = 0.
Step 4, write the tangent line: y − (−2) = 0 · (x − π), which gives y = −2.
The tangent line is the horizontal line y = −2. This makes geometric sense: at x = π, both cos and sec have a local extremum, so the slope is zero.
Students often write (cos x)' = sin x, forgetting the negative sign. The derivative of cosine is negative sine.
In the quotient rule, students swap the order of the subtraction in the numerator. It is f'g − fg', not fg' − f'g. Getting this backwards flips the sign of the entire answer.
When differentiating a product of three functions (like −3x sin x cos x), students sometimes only apply the product rule once and lose a term. You need to account for the derivative of each factor in turn.
Students sometimes simplify before differentiating when they should not, or fail to simplify when it would make the problem far easier (as in Example (d), where rewriting first avoids a messy quotient rule).
⚠️ "Derive the derivative of tan(x) using the quotient rule" is a classic show-your-work problem. Know this proof cold.
⚠️ Product rule problems with trig functions appear on virtually every Calculus I exam. Expect at least two.
⚠️ Tangent line problems combine evaluation, differentiation and point-slope form. Each step is a place to lose marks if you rush.
⚠️ Watch for problems where simplifying the function first (rewriting sec as 1/cos, or tan as sin/cos) makes differentiation dramatically simpler.
Fill in the blank: (cos x)' = ___.
True or False: (sec x)' = sec(x) · cos(x).
Fill in the blank: The quotient rule states (f/g)' = ___.
True or False: The tangent line to f at x = a passes through the point (a, f'(a)).
Fill in the blank: To differentiate x² sin(x), you use the ___ rule.
Answers: 1. −sin x. 2. False (it is sec x tan x). 3. (f'g − fg')/g². 4. False (it passes through (a, f(a))). 5. Product.
Q: Differentiate f(x) = x³ tan(x).
A: Product rule: f'(x) = 3x² tan(x) + x³ sec²(x).
Q: Differentiate g(x) = sin(x) / (1 + cos(x)).
A: Quotient rule: g'(x) = [cos(x)(1 + cos x) − sin(x)(−sin x)] / (1 + cos x)² = [cos x + cos²x + sin²x] / (1 + cos x)² = (cos x + 1) / (1 + cos x)² = 1 / (1 + cos x).
Q: Find the equation of the tangent line to y = sin(x) at x = π/2.
A: f(π/2) = 1, f'(x) = cos(x), f'(π/2) = 0. Tangent line: y = 1.
Q: Use the quotient rule to show that (cot x)' = −csc²(x).
A: Write cot x = cos x / sin x. Apply the quotient rule: [−sin(x) · sin(x) − cos(x) · cos(x)] / sin²(x) = −[sin²x + cos²x] / sin²x = −1/sin²x = −csc²x.
Q: Differentiate h(t) = sec(t) / t².
A: Quotient rule: h'(t) = [sec(t)tan(t) · t² − sec(t) · 2t] / t⁴ = sec(t)[t tan(t) − 2] / t³.
These derivative formulas are the building blocks for the chain rule (Chapter 3), where you will differentiate compositions like sin(x²) or cos(eˣ). The tangent line concept reappears in linear approximation and differentials later in the course.
The quotient rule derivation of (tan x)' = sec²x also previews the method you will use to derive the derivatives of the other four trig functions, all of which can be obtained by writing them in terms of sin and cos.
Related Terms / Search Tags
trig derivatives, derivative of sine, derivative of cosine, derivative of tangent, product rule trig, quotient rule trig, sec squared, tangent line equation, point-slope form, MTH 132, Calculus I, Michigan State University, Section 2.4, differentiation rules