Applications of differentiation

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Turning points and the second derivative

At a turning point the gradient is zero: $\dfrac{dy}{dx} = 0$ .
The second derivative $\dfrac{d^2y}{dx^2}$ $\frac{d ^{2} y}{d x ^{2}}$ (differentiate again) classifies it:
- $\dfrac{d^2y}{dx^2} < 0$ → maximum (curve bends downward).
- $\dfrac{d^2y}{dx^2} > 0$ → minimum (curve bends upward).

Points of inflection

A point of inflection is where the curve changes the way it bends (its concavity).
It occurs where $\dfrac{d^2y}{dx^2} = 0$ and the second derivative changes sign.

Optimisation and related rates

Optimisation: write the quantity in terms of one variable, differentiate, set to zero, solve, and reject impossible answers.
Related rates: two changing quantities are linked, and the chain rule connects their rates, e.g. $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ .

Worked ExampleClassifying turning points

Find and classify the turning points of $y = x^3 - 3x$ .

Step 1 — Set the first derivative to zero

$\frac{dy}{dx} = 3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1 \text{ or } x = -1$

Step 2 — Find the second derivative

$\frac{d^2y}{dx^2} = 6x$

Step 3 — Test each turning point

At $x = 1$ : $6(1) = 6 > 0$ → minimum. $y = 1 - 3 = -2$ , so $(1, -2)$ .
At $x = -1$ : $6(-1) = -6 < 0$ → maximum. $y = -1 + 3 = 2$ , so $(-1, 2)$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Find the $x$ -coordinate of the turning point of $y = x^2 - 10x + 4$ .

Merit

Find the coordinates of the turning points of $y = x^3 - 12x + 1$ and classify each.

Excellence

A closed cylinder has volume $500$ cm³. Its surface area is $A = 2\pi r^2 + \dfrac{1000}{r}$ . Find the radius that minimises the surface area.