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WAMath MethodsSyllabus dot point

How do the product, quotient and chain rules let us differentiate and apply more complex functions?

Apply the product, quotient and chain rules to differentiate functions, and use the derivative in optimisation, rates of change and curve sketching

WACE Year 12 Mathematics Methods Unit 3 further differentiation: the product, quotient and chain rules combined, optimisation, rates of change and the second derivative for curve sketching, with worked SCSA-style examples.

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  1. What this dot point is asking
  2. The three combining rules
  3. Combining the rules
  4. Rates of change
  5. Optimisation
  6. The second derivative and curve sketching

What this dot point is asking

SCSA Unit 3 extends your differentiation toolkit so you can handle functions built from several pieces multiplied, divided or nested inside one another, and then apply the derivative to real contexts: maxima and minima, rates of change, and sketching. This dot point is examined in both the calculator-free and calculator-assumed sections of the WACE written examination.

The three combining rules

The skill SCSA tests is recognising which rule (often several) a function needs and applying them in the correct order. Always differentiate the outermost structure first, then work inwards.

Standard derivatives you must know

  • ddx(xn)=nxn1\dfrac{d}{dx}(x^n) = n x^{n-1}
  • ddx(sinx)=cosx\dfrac{d}{dx}(\sin x) = \cos x, ddx(cosx)=sinx\dfrac{d}{dx}(\cos x) = -\sin x
  • ddx(ex)=ex\dfrac{d}{dx}(e^{x}) = e^{x}, ddx(lnx)=1x\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}

Combining the rules

A single function can require all three rules. The strategy is to label the structure clearly before differentiating.

Rates of change

The derivative dydx\dfrac{dy}{dx} is the instantaneous rate of change of yy with respect to xx. If a quantity QQ depends on time tt, then dQdt\dfrac{dQ}{dt} is the rate of change of QQ. For example, if the volume of water in a tank is V(t)=50tt2V(t) = 50t - t^2 litres, the inflow rate at time tt is V(t)=502tV'(t) = 50 - 2t litres per unit time, which is zero at t=25t = 25 (the tank stops filling).

Optimisation

To find a maximum or minimum of a quantity in context:

  1. Write the quantity to be optimised as a function of one variable, using any constraint to eliminate other variables.
  2. Differentiate and solve f(x)=0f'(x) = 0 for the stationary points.
  3. Classify each stationary point (sign of ff' either side, or the sign of ff'').
  4. Check the endpoints of the domain and interpret the answer in context.

The second derivative and curve sketching

The second derivative f(x)f''(x) measures concavity:

  • f(x)>0f''(x) > 0: the curve is concave up (holds water).
  • f(x)<0f''(x) < 0: the curve is concave down.
  • A point of inflection occurs where ff'' changes sign.

The second derivative test classifies a stationary point x=ax = a where f(a)=0f'(a) = 0: if f(a)>0f''(a) > 0 it is a local minimum, and if f(a)<0f''(a) < 0 it is a local maximum. When f(a)=0f''(a) = 0 the test is inconclusive and you must use a sign table for ff'.

A complete curve sketch combines intercepts, stationary points, the nature of each stationary point, points of inflection, and end behaviour.