1.07q - Product and quotient rules

1.07q - Product and quotient rules

Two changing factors need two contributions to their derivative. Learn the product and quotient rules in 1.07q, and decide when algebra gives a shorter route. You will keep domain restrictions even when a fraction simplifies to something familiar.

Why a product has two derivative terms

If you differentiate each factor of x×xx\times x, should you multiply the answers? That would give one, although x2x^2 has derivative 2x2x. The missing information is the size of each unchanged factor.

You already know

Differentiate powers, ln⁡x\ln x, exponentials and radian trigonometric functions. A derivative is an instantaneous rate; two expressions for the same function must give the same derivative on their common domain.

Let u,vu,v be differentiable. Add and subtract u(x+h)v(x)u(x+h)v(x) in the difference quotient:

u(x+h)v(x+h)−u(x)v(x)h=u(x+h)v(x+h)−v(x)h+v(x)u(x+h)−u(x)h\frac{u(x+h)v(x+h)-u(x)v(x)}h =u(x+h)\frac{v(x+h)-v(x)}h+v(x)\frac{u(x+h)-u(x)}h

As h→0h\to0, differentiability gives continuity of uu, so the limit is uv′+vu′uv'+vu'. Both changes contribute; the tiny product of changes contributes nothing extra in the limit.

Product rule

For differentiable u(x),v(x)u(x),v(x),

ddx(uv)=udvdx+vdudx\frac{\mathrm d}{\mathrm dx}(uv)=u\frac{\mathrm dv}{\mathrm dx}+v\frac{\mathrm du}{\mathrm dx}

Not in the formula sheet: learn it.

A logarithm makes the domain worth writing down before you start.

Worked example 1: a linear factor and a logarithm

Question. Differentiate y=xln⁡xy=x\ln x, stating its domain.

Plan. Differentiate one factor at a time and retain the other.

x>0logarithm domainy′=x(1x)+(ln⁡x)(1)product rule=1+ln⁡xsimplify\begin{aligned} x&>0 &&\text{logarithm domain}\\ y'&=x\left(\frac1x\right)+(\ln x)(1) &&\text{product rule}\\ &=1+\ln x &&\text{simplify} \end{aligned}

Why this step? The retained factors describe how strongly each change affects the whole product.

Answer. y′=1+ln⁡xy'=1+\ln x for x>0x>0.

Check. At one, the function is zero but its gradient is one: height and gradient are different quantities.

Change only the power of the algebraic factor, from one to two.

Worked example 2: a quadratic factor

Question. Differentiate y=x2ln⁡xy=x^2\ln x.

Plan. Use the same two terms with the new derivative 2x2x.

y′=x21x+(ln⁡x)(2x)product rule=x+2xln⁡xsimplify\begin{aligned} y'&=x^2\frac1x+(\ln x)(2x) &&\text{product rule}\\ &=x+2x\ln x &&\text{simplify} \end{aligned}

Why this step? Both the retained power and its derivative change when the power changes.

Answer. y′=x+2xln⁡xy'=x+2x\ln x, for x>0x>0.

Check. At one the gradient is still one; the logarithmic term vanishes there.

The false rule (uv)′=u′v′(uv)'=u'v' also fails spectacularly for x×x−1=1x\times x^{-1}=1: it would give −x−2-x^{-2}, although the product is constant wherever defined.

Watch out

Wrong: multiply the two derivatives. Right: add two terms, differentiating exactly one factor in each.

Raise only the power again.

Applying the product rule to exponential and trig factors

Factoring a derivative often reveals more than expanding it. For an exponential product, the exponential never vanishes, so stationary points come from the remaining factors.

Worked example 3: an exponential product

Question. Differentiate f(x)=x2e−xf(x)=x^2\mathrm e^{-x} and find its stationary points.

Plan. Apply the product rule, then factor before solving f′(x)=0f'(x)=0.

f′(x)=2xe−x−x2e−xproduct rule=xe−x(2−x)factorf′(x)=0  ⟺  x=0 or x=2exponential is positive\begin{aligned} f'(x)&=2x\mathrm e^{-x}-x^2\mathrm e^{-x} &&\text{product rule}\\ &=x\mathrm e^{-x}(2-x) &&\text{factor}\\ f'(x)=0&\iff x=0\text{ or }x=2 &&\text{exponential is positive} \end{aligned}

Why this step? Cancelling an exponential is safe here; cancelling xx would lose a stationary point.

Answer. The stationary points are (0,0)(0,0) and (2,4e−2)(2,4\mathrm e^{-2}).

Check. The derivative has signs negative, positive, negative on the intervals separated by zero and two, giving a minimum then a maximum.

Change only the exponential's input rate from minus one to minus two.

Worked example 4: faster decay

Question. Differentiate g(x)=x2e−2xg(x)=x^2\mathrm e^{-2x} and find its stationary points.

Plan. Keep both product terms and factor their common exponential.

g′(x)=2xe−2x−2x2e−2xinput factor −2=2xe−2x(1−x)factorg′(x)=0  ⟺  x=0 or x=1positive exponential\begin{aligned} g'(x)&=2x\mathrm e^{-2x}-2x^2\mathrm e^{-2x} &&\text{input factor }-2\\ &=2x\mathrm e^{-2x}(1-x) &&\text{factor}\\ g'(x)=0&\iff x=0\text{ or }x=1 &&\text{positive exponential} \end{aligned}

Why this step? Faster decay changes the balance between polynomial growth and exponential decrease.

Answer. The stationary points are (0,0)(0,0) and (1,e−2)(1,\mathrm e^{-2}).

Check. At zero, g′(0)=0g'(0)=0; at one, its two product terms cancel exactly.

The rule handles trigonometric factors too. In radians, (xcos⁡2x)′=cos⁡2x−2xsin⁡2x(x\cos2x)'=\cos2x-2x\sin2x, which has value −π/2-\pi/2 at x=π/4x=\pi/4 and one at zero. For three factors, group two together and apply the rule again; expansion may be shorter for polynomial products.

Exam tip

Show both product terms before factorising. If one standard derivative slips, the correct product structure can still earn a method mark.

Change the same decay rate once more.

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