Lesson 9: Trig Derivatives

(a) Opening Question

A pendulum's horizontal displacement from rest is modeled by s(x) = sin x, where x is measured in radians. We already know from Lesson 1 that the derivative s′(x) measures the instantaneous rate at which displacement changes.

Here is a question you can attempt right now using only the limit definition and a table. Fill in this table of slopes of sin x (use radian mode):

Now compare each slope to the value of cos x at the same point. What do you notice? At x = 0, the slope of sin x is 1, and cos 0 = 1. At x = π/2, the slope is 0, and cos(π/2) = 0.

By the end of this lesson you will be able to say, with proof, exactly what you're noticing: the derivative of sin x is cos x. Keep radians in mind — the whole result depends on it.

(b) Core Concepts

The six rules

These six derivatives are worth memorizing cold. They appear constantly on the AP exam, almost always non-calculator, and the chain-rule, product, and quotient versions all build on them.

A memory device: the "co-" functions get the minus sign

Notice the pattern in the negative signs. The three "co-" functions — cosine, cosecant, cotangent — are exactly the three whose derivatives carry a negative sign:

sin x →  cos x          cos x  → −sin x
tan x →  sec² x         cot x  → −csc² x
sec x →  sec x tan x    csc x  → −csc x cot x

Read down the columns: the left column (sin, tan, sec) has all-positive derivatives; the right column (cos, cot, csc — the "co-" functions) has all-negative derivatives. Also notice the right column is the mirror image of the left: replace every function with its "co-" partner and stick a minus sign on the front. If you memorize the left three and the "co-flip-plus-minus" rule, you get the other three for free.

Where d/dx[sin x] = cos x comes from

This is not magic — it comes straight from the limit definition of the derivative (Lesson 6) plus the two special limits we proved in Lesson 3:

lim_{h→0} (sin h)/h = 1        lim_{h→0} (1 − cos h)/h = 0

Start with the definition:

d/dx[sin x] = lim_{h→0} [sin(x + h) − sin x] / h

Apply the angle-addition identity sin(x + h) = sin x cos h + cos x sin h:

= lim_{h→0} [sin x cos h + cos x sin h − sin x] / h
= lim_{h→0} [sin x (cos h − 1) + cos x sin h] / h

Split the fraction and pull the constants sin x and cos x (constant with respect to h) outside their limits:

= sin x · lim_{h→0} (cos h − 1)/h  +  cos x · lim_{h→0} (sin h)/h
= sin x · (0)  +  cos x · (1)
= cos x

That's the whole derivation. The two special limits do all the heavy lifting — which is exactly why those limits matter and why they live in the limits unit. The derivative of cos x follows from the identical argument with cos(x + h) = cos x cos h − sin x sin h, producing the −sin x.

Where the other four come from (quotient rule)

You don't have to memorize the tan/sec/csc/cot rules as separate facts if you remember that they're all sin and cos in disguise. For example, tan x = sin x / cos x, so by the quotient rule:

d/dx[tan x] = [cos x · cos x − sin x · (−sin x)] / cos² x
            = (cos² x + sin² x) / cos² x
            = 1 / cos² x
            = sec² x

The same trick (writing sec x = 1/cos x, cot x = cos x/sin x, csc x = 1/sin x) recovers all four. On the exam, recall the rule directly — but knowing the derivation means you can rebuild any rule you blank on.

Radians are non-negotiable

Every rule above assumes x is in radians. The derivation used lim_{h→0}(sin h)/h = 1, which is true only in radians (in degrees that limit is π/180). If your calculator is in degree mode, nDeriv of sin x will be wrong by a factor of π/180 ≈ 0.01745. Set radian mode for any numerical check this unit.

Combining with product and quotient rules

Trig derivatives slot directly into the rules from Lesson 8. A quick worked example combining the quotient rule with trig:

Differentiate f(x) = sin x / (1 + cos x).

f′(x) = [(1 + cos x)·(cos x) − (sin x)·(−sin x)] / (1 + cos x)²
      = [cos x + cos² x + sin² x] / (1 + cos x)²
      = [cos x + 1] / (1 + cos x)²        (since cos² x + sin² x = 1)
      = 1 / (1 + cos x)

The Pythagorean identity collapsed the whole numerator. Watch for these simplifications — the AP answer key often expects the reduced form.

(c) Worked Examples

Example 1 — Product rule with trig (foundation) · [NO CALC]

Problem: Find f′(x) for f(x) = x² sin x.

Strategy: This is a product of x² and sin x, so use the product rule (uv)′ = u′v + uv′ with u = x², v = sin x.

Solution:

u = x²        → u′ = 2x
v = sin x     → v′ = cos x

f′(x) = u′v + uv′
      = (2x)(sin x) + (x²)(cos x)
      = 2x sin x + x² cos x

Justification: The derivative of sin x is cos x by the trig rule; the product rule was required because f is a product of two non-constant functions of x.

Example 2 — Quotient rule with trig (procedural) · [NO CALC]

Problem: Find g′(x) for g(x) = (sec x) / x.

Strategy: Quotient rule (u/v)′ = (u′v − uv′)/v² with u = sec x, v = x.

Solution:

u = sec x     → u′ = sec x tan x
v = x         → v′ = 1

g′(x) = [ (sec x tan x)(x) − (sec x)(1) ] / x²
      = [ x sec x tan x − sec x ] / x²
      = sec x (x tan x − 1) / x²

Justification: d/dx[sec x] = sec x tan x (not sec² x — that's the derivative of tan x). The quotient rule applies because g is a ratio with the variable in the denominator.

Example 3 — Evaluating a trig derivative at a special angle · [NO CALC]

Problem: For h(x) = 2 sin x − cos x, find h′(π/6).

Strategy: Differentiate term by term, then substitute the special angle. Recall cos(π/6) = √3/2 and sin(π/6) = 1/2.

Solution:

h′(x) = 2 cos x − (−sin x) = 2 cos x + sin x

h′(π/6) = 2·cos(π/6) + sin(π/6)
        = 2·(√3/2) + (1/2)
        = √3 + 1/2

So h′(π/6) = √3 + 1/2 ≈ 2.232.

Justification: The −cos x term differentiates to −(−sin x) = +sin x — the double negative is the single most common sign slip here.

Example 4 — Tangent line to a trig curve (AP level) · [NO CALC]

Problem: Write the equation of the line tangent to y = tan x at x = π/4.

Strategy: A tangent line needs a point and a slope. The point is (π/4, tan(π/4)); the slope is y′(π/4).

Solution:

Point:  tan(π/4) = 1,  so the point is (π/4, 1)

Slope:  y′ = sec² x
        y′(π/4) = sec²(π/4) = (√2)² = 2

Tangent line (point-slope):
        y − 1 = 2(x − π/4)
        y = 2x − π/2 + 1

Justification: sec(π/4) = 1/cos(π/4) = 1/(√2/2) = √2, so sec²(π/4) = 2. The derivative sec² x gives the slope at any point; evaluating at π/4 gives the slope of this particular tangent line.

(d) Common Mistakes

Wrong sign on d/dx[cos x].

Students write d/dx[cos x] = sin x. Why it's wrong: cosine is a "co-" function, so its derivative is negative: d/dx[cos x] = −sin x. Fix: Memorize the "co-functions get the minus sign" rule and double-check every cosine.

Confusing sec² x with sec x tan x.

These look similar and get swapped constantly. Why it's wrong: d/dx[tan x] = sec² x, but d/dx[sec x] = sec x tan x. Fix: Tangent → secant squared; secant → secant times tangent. Memorize the short phrase: tan gives sec², sec gives sec·tan.

Degree mode instead of radian mode.

A student checks f′ with nDeriv in degree mode and gets a number off by ≈ 0.01745. Why it's wrong: the rules d/dx[sin x] = cos x etc. are only valid in radians. Fix: Set the calculator to RADIAN (MODE → Radian) before any trig derivative check.

Forgetting the product/quotient rule with trig.

For x² sin x, students write 2x cos x. Why it's wrong: they differentiated each factor separately — that is not how products work. Fix: x² sin x is a product, so use the product rule: 2x sin x + x² cos x.

Treating sin² x as a basic trig derivative.

Note sin² x means (sin x)², a composite — that needs the chain rule (Lesson 12), not a rule from this lesson. Fix: In this lesson keep arguments and powers plain; don't try to differentiate sin²x or sin(3x) yet.

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## (e) Practice Problems

Short answer. Find f′(x) for f(x) = x² tan x. Show the rule you used.

Short answer. Find the equation of the line tangent to y = sin x at x = 0.

Short answer. Differentiate r(x) = (cos x)/x².

Justification. A student claims d/dx[cot x] = csc² x. Is the student correct? Justify your answer, including the correct derivative.

Justification. For f(x) = x sin x, a classmate computes f′(π) = π. Justify whether this is correct by showing the computation, and state the value of f′(π).

Short answer. Evaluate d/dx[2 sin x + 3 cos x] at x = π.

The height (in meters) of a weight bobbing on a spring is H(t) = 4 + cos t for t in seconds. Find the rate of change of height at t = 1 second (radians), to three decimal places.

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## (f) AP Exam Focus — Free Response

> FRQ 9 (9 points total). No calculator.

>

> A buoy floats on the ocean surface. Its height above mean sea level, in feet, is modeled by

>

> `

> H(t) = 6 + 2 sin t

> `

>

> where t is measured in radians and 0 ≤ t ≤ 2π. (Here t is a normalized phase variable, not seconds.)

>

> (a) (2 points) Find H′(t). Find the rate of change of the buoy's height at t = π/3, and indicate units.

>

> (b) (3 points) Write an equation for the line tangent to the graph of H at t = π/3. Use it to approximate the buoy's height at t = π/3 + 0.1.

>

> (c) (2 points) Find the value of t in (0, 2π) at which the buoy reaches its maximum height, and justify that it is a maximum.

>

> (d) (2 points) The buoy's height is increasing on certain intervals. Using H′(t), interpret what H′(π/2) = 0 tells you about the motion of the buoy at that instant.

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### Model Solution

(a) Differentiate term by term. The derivative of the constant 6 is 0, and d/dx[2 sin t] = 2 cos t:

H′(t) = 2 cos t

At t = π/3:

H′(π/3) = 2 cos(π/3) = 2·(1/2) = 1

The rate of change of the buoy's height at t = π/3 is 1 foot per radian (height units per unit of t).

(b) The tangent line needs the point (π/3, H(π/3)) and the slope H′(π/3).

H(π/3) = 6 + 2 sin(π/3) = 6 + 2·(√3/2) = 6 + √3
H′(π/3) = 1   (from part a)

Tangent line (point-slope form):

y − (6 + √3) = 1·(t − π/3)
y = t − π/3 + 6 + √3

Approximating at t = π/3 + 0.1 (so t − π/3 = 0.1):

y ≈ 0.1 + 6 + √3 = 6.1 + √3 ≈ 7.832 feet

(c) The maximum of H(t) = 6 + 2 sin t occurs where sin t is largest. Find critical points by setting H′(t) = 0:

H′(t) = 2 cos t = 0   ⟹   t = π/2  or  t = 3π/2   in (0, 2π)

At t = π/2: H′ changes from positive (H′(t) > 0 for t just below π/2, since cos t > 0) to negative (H′(t) < 0 just above π/2, since cos t < 0).

By the First Derivative Test, H has a relative maximum at t = π/2 because H′ changes from positive to negative there. The maximum height is H(π/2) = 6 + 2 sin(π/2) = 6 + 2 = 8 feet, occurring at t = π/2.

(d) H′(π/2) = 2 cos(π/2) = 0. Since H′(t) is the instantaneous rate of change of the buoy's height with respect to t, H′(π/2) = 0 means that at t = π/2 the buoy's height is instantaneously neither rising nor falling — its vertical velocity is zero. This is the instant the buoy reaches the top of its motion (the crest), where it momentarily stops before beginning to descend.

Scoring Commentary

Part (a) — 2 points. 1 point for the correct derivative H′(t) = 2 cos t; 1 point for H′(π/3) = 1 with units. Where students lose points: writing H′(t) = 2 sin t (forgetting the derivative of sin is cos), or omitting units.

Part (b) — 3 points. 1 point for the point (π/3, 6 + √3); 1 point for using slope 1 to write a correct tangent-line equation; 1 point for the approximation ≈ 6.1 + √3. Where students lose points: a slope or point error propagates, but the approximation point can still be earned if their tangent line is used consistently.

Part (c) — 2 points. 1 point for finding t = π/2 from H′(t) = 0; 1 point for the justification that H′ changes sign positive-to-negative (a First Derivative Test argument). Where students lose points: stating "it's a max" without sign analysis earns the point-finding but not the justification point. Listing both π/2 and 3π/2 without selecting the maximum loses the answer point.

Part (d) — 2 points. 1 point for H′(π/2) = 0; 1 point for the interpretation in context — that the height is momentarily not changing / vertical velocity is zero / the buoy is at the crest. Where students lose points: a bare "it equals zero" with no contextual meaning earns only the computation point. The AP rubric requires interpreting the value as a rate in the buoy context.

f′(x) = 2x tan x + x² sec² x

r′(x) = [(−sin x)(x²) − (cos x)(2x)] / (x²)²
      = [−x² sin x − 2x cos x] / x⁴
      = (−x sin x − 2 cos x) / x³        (factoring x from numerator and reducing)