Difficulty: Intermediate | Prerequisites: Product rule, quotient rule, basic trig identities, Section 2.3 notes.
Differentiating trigonometric functions means applying the product rule and quotient rule alongside the two base derivatives: (sin x)′ = cos x and (cos x)′ = −sin x. From those two, every other trig derivative (tan, sec, csc, cot) can be derived. The section also covers tangent line equations, which combine evaluation with differentiation.
Product rule
If f(x) = g(x) · h(x), then f′(x) = g′(x) · h(x) + g(x) · h′(x). Think of it as: differentiate the first and leave the second alone, then leave the first alone and differentiate the second, then add.
Quotient rule
If f(x) = g(x)/h(x), then f′(x) = [g′(x) · h(x) − g(x) · h′(x)] / [h(x)]². Think of it as: "low d-high minus high d-low, over the square of what's below."
Tangent line
The line that touches a curve at exactly one point and has slope equal to the derivative at that point. The equation is y − f(a) = f′(a)(x − a).
sec(x)
sec(x) = 1/cos(x). Its derivative is sec(x) tan(x). This comes up frequently when differentiating tan(x) or expressions involving 1/cos(x).
When a function is the product of two pieces (one or both involving trig), apply the product rule. For products of three factors, apply the rule iteratively or use the extended product rule.
Example (a): f(x) = ³√x · sec(x) = x^(1/3) sec(x)
f′(x) = (1/3)x^(−2/3) sec(x) + x^(1/3) sec(x) tan(x)
The first term differentiates the power; the second differentiates the sec.
Example (b): g(x) = x² cos(x) − x^(1/4)
g′(x) = 2x cos(x) + x²(−sin(x)) − (1/4)x^(−3/4)
The product rule applies only to x² cos(x); the −x^(1/4) term differentiates separately.
Example (e): f(x) = −3x sin(x) cos(x)
This is a product of three factors: (−3x), sin(x), cos(x).
f′(x) = −3 sin(x) cos(x) − 3x cos(x) cos(x) − 3x sin(x)(−sin(x))
Simplifies to: −3 sin(x) cos(x) − 3x cos²(x) + 3x sin²(x)
Use the quotient rule whenever a trig function sits in the denominator.
Example (c): h(x) = x³ / (10 + tan(x))
h′(x) = [3x²(10 + tan(x)) − x³ sec²(x)] / (10 + tan(x))²
The numerator follows "low d-high minus high d-low" exactly.
Example (d): p(t) = (2 tan(t) + 2) / sec(t)
p′(t) = [2 sec²(t) · sec(t) − (2 tan(t) + 2) · sec(t) tan(t)] / sec²(t)
Problem 6 asks you to prove (tan x)′ = sec²(x) using the quotient rule. This is a common exam proof.
Write tan(x) = sin(x)/cos(x)
Apply the quotient rule: [cos(x) · cos(x) − sin(x) · (−sin(x))] / cos²(x)
Numerator simplifies: cos²(x) + sin²(x) = 1
Result: 1/cos²(x) = sec²(x)
To find the equation of the tangent line at x = a:
Compute f(a) to find the y-coordinate of the point.
Compute f′(a) to find the slope.
Write the line: y − f(a) = f′(a)(x − a).
Example: f(x) = cos(x) + sec(x) at x = π
f(π) = cos(π) + sec(π) = −1 + (−1) = −2
f′(x) = −sin(x) + sec(x) tan(x)
f′(π) = −sin(π) + sec(π) tan(π) = 0 + 0 = 0
Tangent line: y + 2 = 0(x − π), which gives y = −2
A horizontal tangent line, which means the curve has a local extremum at x = π.
\frac{d}{dx}[\sin x] = \cos x\frac{d}{dx}[\cos x] = -\sin x\frac{d}{dx}[\tan x] = \sec^2 x\frac{d}{dx}[\sec x] = \sec x \tan xProduct rule:
(fg)' = f'g + fg'Quotient rule:
\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}Tangent line at x = a:
y - f(a) = f'(a)(x - a)Derivatives of trig functions describe rates of change in oscillating systems. The velocity of a weight on a spring, the current in an AC circuit, and the rate of change of daylight hours through the year are all modelled by derivatives of sine and cosine.
Students often write (cos x)′ = sin x, forgetting the negative sign. The derivative of cosine is negative sine.
When applying the product rule to three factors, students frequently differentiate only two of them. Each factor gets its turn being differentiated while the other two are left alone, giving three terms.
Students sometimes apply the quotient rule but reverse the order in the numerator. The correct order is (bottom × derivative of top) minus (top × derivative of bottom), not the other way round.
When finding tangent lines, students compute f′(a) but forget to also evaluate f(a) for the point itself.
⚠️ Expect at least one product rule problem and one quotient rule problem involving trig on Exam 2.
⚠️ The proof that (tan x)′ = sec²(x) is a classic "show your work" question. Know it cold.
⚠️ Tangent line problems combine two skills (evaluate and differentiate), so they are a favourite for testing overall fluency.
⚠️ Watch for problems where the slope turns out to be zero (horizontal tangent). This is not an error; it means the curve is flat at that point.
True or false: The derivative of cos(x) is sin(x).
False. It is −sin(x).
Fill in the blank: The quotient rule numerator is (low × d-high) ___ (high × d-low).
Minus.
True or false: sec²(x) = 1/cos²(x).
True.
Fill in the blank: The tangent line to f at x = a has slope ___.
f′(a).
True or false: If f′(a) = 0, there is no tangent line at x = a.
False. The tangent line is horizontal (slope zero).
Q: Find f′(x) if f(x) = ³√x sec(x).
A: Write f(x) = x^(1/3) sec(x). By the product rule: f′(x) = (1/3)x^(−2/3) sec(x) + x^(1/3) sec(x) tan(x).
Q: Differentiate h(x) = x³ / (10 + tan(x)).
A: By the quotient rule: h′(x) = [3x²(10 + tan(x)) − x³ sec²(x)] / (10 + tan(x))².
Q: Use the quotient rule to show that (tan x)′ = sec²(x).
A: tan(x) = sin(x)/cos(x). Quotient rule gives [cos(x) cos(x) − sin(x)(−sin(x))] / cos²(x) = [cos²(x) + sin²(x)] / cos²(x) = 1/cos²(x) = sec²(x).
Q: Find the tangent line to f(x) = cos(x) + sec(x) at x = π.
A: f(π) = −1 + (−1) = −2. f′(π) = −sin(π) + sec(π)tan(π) = 0. The tangent line is y = −2.
Q: Differentiate f(x) = −3x sin(x) cos(x).
A: Apply the product rule to three factors: f′(x) = −3 sin(x) cos(x) − 3x cos²(x) + 3x sin²(x).
These derivative techniques lead directly into the chain rule (next section), where you will differentiate compositions like sin(x²) or cos(3x + 1). The tangent line equation also reappears in linear approximation and differentials later in the course.
derivatives of trig functions, product rule, quotient rule, tangent line equation, derivative of sin, derivative of cos, derivative of tan, sec squared, proof tan derivative, MTH 132, Calculus I, Michigan State, Section 2.4, differentiation rules, slope of tangent