Understand · explore · practise
Understand and apply the product rule with a changing-area explorer, original worked examples and independent practice. Combine it with the chain rule, factorise derivatives and solve tangent and stationary-point problems.
Before you startBasic derivatives; chain rule; factorising expressions
01 / Account for both changing factors
If y = uv, then dy/dx = u(dv/dx) + v(du/dx).
Both u and v are functions of x and must be differentiable at the input being considered.
The product can change because u changes, because v changes, or because both do. Keeping only one of those contributions misses part of the derivative.
In the explorer, the original rectangle has area uv. The blue and green strips give the two first-order changes. The gold corner is a smaller second-order change.
x = 1; u = 3; v = 5. Right strip = 2.5; top strip = 3; corner = 0.5.
Exact area change = 6; average rate = 12. Derivative at x = 1 is 11.
The corner contributes 2h to the average rate. This tends to zero as h shrinks; the two strip contributions remain.
The picture uses positive lengths and positive steps. The algebraic product rule also works for negative function values and rates, wherever the factors are differentiable.
02 / Explain the two terms
(u + Δu)(v + Δv) − uv = u Δv + v Δu + Δu Δv
Subtract the original rectangle.
Δ(uv)/h = u(Δv/h) + v(Δu/h) + (Δu/h)Δv
Divide every contribution by the input step.
As h → 0: (uv)′ = uv′ + vu′
Differentiability makes Δu/h finite in the limit and Δv tend to zero.
In the model, u = x + 2 and v = 2x + 3. The derivative is 2u + v = 4x + 7. The finite average rate is 4x + 7 + 2h, so it approaches the derivative as h shrinks.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Why can the derivative of uv not generally be u′v′? Test u = x and v = x.
The original product is x².
The derivative of x² is 2x. Multiplying the derivatives would give 1 × 1 = 1, which is not the derivative. The correct sum is x·1 + x·1 = 2x.
03 / Keep a tidy two-factor layout
u = x² + 3, u′ = 2x; v = 2x − 1, v′ = 2
Write each factor beside its derivative.
y′ = 2(x² + 3) + 2x(2x − 1)
Each term contains one unchanged factor.
y′ = 6x² − 2x + 6
Expanding the original first gives the same result.
Differentiate (x² − 2)(3x + 4).
Use 3(x² − 2) + 2x(3x + 4).
The derivative is 9x² + 8x − 6.
Differentiate 7(x² + 1). Does it require a full product-rule calculation?
The derivative of the constant 7 is zero.
The derivative is 14x. The constant-multiple rule is sufficient; the product rule gives 7·2x + (x² + 1)·0.
04 / Combine product and chain rules
u′ = 2x; v′ = 3e³ˣ
The exponential derivative includes the inner factor 3.
y′ = 3(x² + 1)e³ˣ + 2xe³ˣ
Use both contributions.
y′ = e³ˣ(3x² + 2x + 3)
Factor out the common exponential.
Differentiate x²e⁻²ˣ and factorise the answer.
The exponential derivative is −2e⁻²ˣ.
y′ = 2xe⁻²ˣ − 2x²e⁻²ˣ = 2x(1 − x)e⁻²ˣ.
Differentiate (2x + 1)eˣ.
The linear factor has derivative 2.
y′ = (2x + 1)eˣ + 2eˣ = (2x + 3)eˣ.
05 / Include both the product and angle changes
u′ = 1; v′ = 2cos(2x)
Angles are in radians.
y′ = 2x cos(2x) + sin(2x)
The x factor stays in the term where sine is differentiated.
Differentiate x²cos(3x).
The cosine derivative is −3sin(3x).
y′ = 2x cos(3x) − 3x²sin(3x).
Differentiate sin x cos x, then simplify.
Keep the minus sign from differentiating cosine.
y′ = cos²x − sin²x = cos(2x). Differentiating (1/2)sin(2x) gives the same result.
06 / Retain the real logarithm domain
Domain: x > 0
The logarithm must exist.
y′ = x²(1/x) + 2x ln x
Differentiate one factor at a time.
y′ = x(1 + 2ln x), x > 0
Keep the domain after simplifying.
Differentiate (x + 2)ln x and state the domain.
Use (x + 2)/x + ln x.
y′ = 1 + 2/x + ln x, for x > 0.
Differentiate x ln(x² + 1).
The logarithm derivative is 2x/(x² + 1).
y′ = ln(x² + 1) + 2x²/(x² + 1), valid for every real x because x² + 1 is positive.
07 / Differentiate powers inside each factor
u′ = 2(x + 1); v′ = 6(2x − 3)²
Each factor uses the chain rule.
y′ = 6(x + 1)²(2x − 3)² + 2(x + 1)(2x − 3)³
Use the product rule outside.
y′ = 2(x + 1)(2x − 3)²[3(x + 1) + (2x − 3)]
Extract the smallest common powers.
y′ = 10x(x + 1)(2x − 3)²
The remaining bracket simplifies to 5x.
Differentiate x√(x + 4), stating where the formula is valid.
The original domain is x ≥ −4; differentiate at interior points.
y′ = √(x + 4) + x/[2√(x + 4)] = (3x + 8)/[2√(x + 4)], valid for x > −4. There is no finite derivative at the endpoint.
08 / Extend to three factors
(uvw)′ = u′vw + uv′w + uvw′
u′ = 1; v′ = eˣ; w′ = cos x
There are three changing factors.
y′ = eˣsin x + xeˣsin x + xeˣcos x
Include one term for each derivative.
y′ = eˣ[(1 + x)sin x + x cos x]
Factor to make the structure easier to read.
You can derive the three-factor rule by applying the ordinary product rule to (uv)w and then to uv.
Differentiate x(x + 1)(x − 2).
Either use three terms or expand first.
y′ = (x + 1)(x − 2) + x(x − 2) + x(x + 1) = 3x² − 2x − 2.
09 / Evaluate a product derivative exactly
y = (π/4)sin(π/2) = π/4
The point is (π/4, π/4).
y′ = sin(2x) + 2x cos(2x); y′(π/4) = 1
Use exact trigonometric values.
y − π/4 = x − π/4, so y = x
A tangent may have a simpler form than the curve.
Find the normal to y = (x + 1)eˣ at x = 0.
The point is (0, 1) and the derivative is (x + 2)eˣ.
The tangent slope is 2, so the normal slope is −1/2. Its equation is y = 1 − x/2.
10 / Factor before solving y′ = 0
y′ = e⁻ˣ − (x − 2)e⁻ˣ = (3 − x)e⁻ˣ
Keep the negative sign from the exponential derivative.
e⁻ˣ > 0, so y′ = 0 only when x = 3
The exponential never equals zero.
The point is (3, e⁻³)
Return to the original function.
y′ is positive before 3 and negative after 3
This point is a maximum.
Find and classify the stationary points of y = x²e⁻ˣ.
Factor y′ = x(2 − x)e⁻ˣ and use a sign table.
The points are (0, 0) and (2, 4e⁻²). The derivative is negative for x < 0, positive for 0 < x < 2, and negative for x > 2. Hence the origin is a minimum and the other point is a maximum.
Find and classify the stationary point of y = x ln x.
The domain is x > 0 and y′ = ln x + 1.
x = e⁻¹ and y = −e⁻¹. Since y″ = 1/x > 0, this is a minimum.
11 / Choose the rule from the outside operation
Section 1 of 11 · Account for both changing factors