A. Rate of Change · Derivatives of Exponential and Sinusoidal Functions

The Chain Rule with Sinusoidal Functions

A3.5About 70 minutes

Learning goals

I can differentiate sin(g(x)) and cos(g(x)) using the chain rule.
I can differentiate transformed sinusoidal functions y = a sin(k(x − d)) + c and explain how a and k change the rates.
I can tell apart and differentiate powers of trigonometric functions (sin² x) and trigonometric functions of powers (sin(x²)).
I can combine sinusoidal and exponential functions in compositions such as e^(sin x) and cos(eˣ).

Words to know

transformed sinusoidal function
y = a sin(k(x − d)) + c or y = a cos(k(x − d)) + c, with amplitude |a|, period 2π/|k|, phase shift d and vertical shift c.

amplitude
Half the distance from the maximum to the minimum of a sinusoidal function: |a|.

period
The horizontal length of one complete cycle: 2π/|k| for y = a sin(k(x − d)) + c.

sin² x
Short for (sin x)², the square of sin x. It is not sin(x²).

composite function
A function made by putting one function inside another, f(g(x)).

Real waves are transformed waves

The height of a tide, the hours of daylight in Thunder Bay, the height of a Ferris wheel seat: these are all sinusoidal, but rarely plain sin x. They look like y = a sin(k(x − d)) + c, with stretches, compressions and shifts that you studied in MHF4U Radian Measure and Trigonometric Functions.

To differentiate them, use the chain rule from Derivatives and Differentiation Rules with the two rules from the last two lessons.

Chain rule versions for sine and cosine

d/dx sin(g(x)) = cos(g(x)) × g′(x)
d/dx cos(g(x)) = −sin(g(x)) × g′(x)

Differentiate the outer trig function, keep the inside exactly as it is, then multiply by the derivative of the inside.

The simplest case: sin(kx)

The inner function kx has derivative k, so d/dx sin(kx) = k cos(kx) and d/dx cos(kx) = −k sin(kx).

A graph explains the extra factor k. The graph of y = sin(2x) is y = sin x compressed horizontally by a factor of 1/2. Each rise happens over half the run, so every slope doubles.

y = sin 2x completes two cycles in the time sin x completes one. At x = 0 its tangent has slope 2, twice the slope of sin x.

The general transformed wave

For y = a sin(k(x − d)) + c:

y′ = a cos(k(x − d)) × k = ak cos(k(x − d)).

The vertical shift c disappears (the derivative of a constant is 0): shifting a wave up does not change its slopes.
The phase shift d stays inside: the derivative wave is shifted by the same amount.
The amplitude a and the factor k multiply. So the greatest rate of change is |ak|. You estimated this in MHF4U Rates of Change; now you can prove it, since the largest value of |cos| is 1.

Worked example

Example 1: Inside functions that are linear

Differentiate (a) y = sin(3x) (b) y = cos(2x − π) (c) y = 5 cos(x/4)

(a) Inner 3x, derivative 3: y′ = 3 cos(3x).
(b) Inner 2x − π, derivative 2: y′ = −sin(2x − π) × 2 = −2 sin(2x − π).
(c) Inner x/4, derivative 1/4: y′ = 5 × (−sin(x/4)) × 1/4 = −(5/4) sin(x/4).

(a) 3 cos 3x (b) −2 sin(2x − π) (c) −(5/4) sin(x/4)

Worked example

Example 2: A fully transformed wave

For y = 4 sin(2(x − π/6)) + 1, find y′, the slope at x = π/6, and the greatest rate of change.

y′ = 4 cos(2(x − π/6)) × 2 = 8 cos(2(x − π/6)).
At x = π/6: 8 cos 0 = 8.
|cos| is at most 1, so the greatest rate is |ak| = 4 × 2 = 8. It happens at x = π/6, where the wave crosses its midline y = 1 going up.

y′ = 8 cos(2(x − π/6)); slope 8 at x = π/6, which is also the greatest rate of change.

Powers of trig functions versus trig functions of powers

These two look alike but are very different:

sin² x = (sin x)²: the sine is calculated first, then squared. The outer function is the square.
sin(x²): x is squared first, then the sine is taken. The outer function is sine.

Read the layers from the inside out. In sin² x the outer layer is squaring; in sin(x²) the outer layer is sine. The chain rule handles each layer in turn.

Worked example

Example 3: sin² x and sin(x²)

Differentiate (a) y = sin² x (b) y = sin(x²) (c) y = cos³ x

(a) y = (sin x)². Outer u², inner sin x: y′ = 2 sin x × cos x = 2 sin x cos x. (By the MHF4U double angle formula, this equals sin 2x.)
(b) Outer sin u, inner x²: y′ = cos(x²) × 2x = 2x cos(x²).
(c) y = (cos x)³: y′ = 3(cos x)² × (−sin x) = −3 cos² x sin x.

(a) 2 sin x cos x (b) 2x cos(x²) (c) −3 cos² x sin x

Worked example

Example 4: Mixing exponentials and trig

Differentiate (a) y = e^(sin x) (b) y = cos(eˣ) (c) y = √(1 + sin x)

(a) Outer eᵘ, inner sin x: y′ = e^(sin x) × cos x.
(b) Outer cos u, inner eˣ: y′ = −sin(eˣ) × eˣ.
(c) y = (1 + sin x)^(1/2): y′ = (1/2)(1 + sin x)^(−1/2) × cos x = cos x ÷ (2√(1 + sin x)).

(a) cos x e^(sin x) (b) −eˣ sin(eˣ) (c) cos x / (2√(1 + sin x))

Hands-on activity

Bicycle gears: rates multiply

You need: a bicycle with gears (or a picture of one from a manual or website), a helper to hold it steady, and paper and pencil. Safety: turn the bike upside down on a soft surface, turn the pedals slowly by hand, and keep fingers away from the chain, gears and spokes.

Count the teeth on the front chainring in use (F) and on the rear cog in use (R).
Turn the pedals exactly one full turn and count how many times the back wheel turns. Compare with F ÷ R.
If the pedal angle is θ, the wheel angle is (F/R)θ. A reflector at radius r on the wheel has height h = r sin((F/R)θ) above the axle (starting level with the axle). Use the chain rule to find dh/dθ.
Explain in words why the factor F/R appears in dh/dθ.

What to look for: with F = 44 and R = 22 (a common pairing), the wheel turns twice per pedal turn, matching F/R = 2. The derivative is dh/dθ = r(F/R) cos((F/R)θ). The factor F/R appears because the wheel turns F/R times as fast as the pedals: that is the chain rule. Rates of nested quantities multiply.

Just for fun

A composite function is like a Nanaimo bar: to get to the middle you have to go through every layer, and the chain rule makes sure each layer gets its own derivative.

Your turn

Differentiate y = sin(5x).

Differentiate y = cos(πx).

Differentiate y = 3 sin(2x + 1) − 4.

Differentiate y = sin(x³).

Differentiate y = sin⁴ x.

Differentiate y = e^(cos x).

What is the greatest rate of change of y = 6 cos(3x) + 2?

More practice: warm-up

d/dx sin(2x) = ?

d/dx cos(4x) = ?

Does the vertical shift c affect the derivative of a sin(kx) + c?

Write sin² x without the shorthand.

Identify the inner and outer functions of cos(x² + 1).

More practice: on your own

Find the slope of y = 2 sin(3x) at x = π/9.

Differentiate y = sin²(3x).

Find the equation of the tangent to y = cos(2x) at x = π/4.

Differentiate y = √(cos x) and state where it is defined on [0, 2π].

Find all x in [0, π] where y = sin² x has a horizontal tangent.

Challenge

Differentiate y = sin(cos(x²)).

Show by differentiating that sin² x + cos² x has derivative 0, and explain why that makes sense.

A wave is modelled by y = a sin(kx) with a > 0 and k > 0. Its greatest height is 3 and its greatest slope is 12. Find a, k and the period.

Check your understanding

Quiz

d/dx sin(4x) =

d/dx cos(x²) =

d/dx sin² x =

The greatest rate of change of y = 5 sin(2x) − 3 is:

d/dx e^(sin x) =

Which function has derivative −6 sin(3x)?

For parents and teachers

Big idea (A3.5): the chain rule lets students differentiate any transformed or nested sinusoidal function. For y = a sin(k(x − d)) + c, the derivative is ak cos(k(x − d)), so the greatest rate of change is |ak|.

How to help: before differentiating, ask your teen to name the layers out loud from the outside in (for sin² x: "square, then sine"; for sin(x²): "sine, then square"). Most errors come from peeling the layers in the wrong order.

Common difficulties:

Forgetting the inner derivative (writing d/dx sin 3x = cos 3x).
Confusing sin² x with sin(x²).
Changing the inside: d/dx sin(3x) is 3 cos(3x), not 3 cos x.
Sign errors with cosine in nested functions.