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Mathematics · Ch 6 — Application of Derivatives

Rate of Change of Quantities

6.2

Rate of Change of Quantities

6.2 Rate of Change of Quantities

The Fundamental Idea

When one quantity yy depends on another quantity xx through y=f(x)y = f(x), the derivative dydx\frac{dy}{dx} (or f′(x)f'(x)) gives the instantaneous rate of change of yy with respect to xx — how fast yy is changing at a particular value of xx. Evaluated at x=x0x = x_0:

dydx∣x=x0orf′(x0)\left.\frac{dy}{dx}\right|_{x = x_0} \quad \text{or} \quad f'(x_0)

Note

The derivative dsdt\frac{ds}{dt} from earlier chapters — the rate of change of distance with respect to time — is just one instance; the same idea applies to any two related quantities.

The Chain Rule Connection

Often both xx and yy vary with respect to a third variable, usually time tt: x=f(t)x = f(t) and y=g(t)y = g(t). We cannot write yy directly as a function of xx, but the Chain Rule still gives:

dydx=dydtdxdt,provided dxdt≠0\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \text{provided } \frac{dx}{dt} \neq 0

Important

This follows from the Chain Rule dydt=dydx⋅dxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}, rearranged when dxdt≠0\frac{dx}{dt} \neq 0.

Sign of the Rate of Change

  • Positive dydx\frac{dy}{dx}: yy increases as xx increases.
  • Negative dydx\frac{dy}{dx}: yy decreases as xx increases.

Economic Rates

ConceptFormula
Marginal CostMC=dCdx\text{MC} = \frac{dC}{dx} (rate of change of total cost)
Marginal RevenueMR=dRdx\text{MR} = \frac{dR}{dx} (rate of change of total revenue)
Tip

For related-rates problems:

  1. Identify the given rates and the rate you need.
  2. Write the geometric or economic relationship between the quantities.
  3. Differentiate with respect to time tt (Chain Rule where needed).
  4. Substitute known values and solve for the unknown rate.
  5. Check the sign — decreasing quantities give negative rates.

A Quick Worked Illustration

Suppose a circular ripple spreads outward so its radius rr grows at drdt=3 cm/s\dfrac{dr}{dt} = 3\text{ cm/s}. How fast is the enclosed area A=πr2A = \pi r^2 growing when r=4 cmr = 4\text{ cm}?

Differentiate AA with respect to tt using the Chain Rule (since AA depends on rr, and rr depends on tt): …