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E2.12 Differentiation: The Power Rule, Gradients & Turning Points

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01 — Explanation

What does \dfrac{dy}{dx} actually mean?

A straight line has one fixed gradient everywhere, but a curve's steepness changes from point to point. Differentiation produces a new function, $\dfrac{dy}{dx}$ , that gives the exact gradient of the tangent to the curve at any chosen $x$ -value — replacing the old estimate-by-drawing-a-tangent method with an exact algebraic one.

02 — Explanation

How do I differentiate ax^n?

Multiply by the existing power, then reduce the power by $1$ :

y=ax^n \implies \frac{dy}{dx}=nax^{n-1}

Term	Derivative	Note
$x^n$	$nx^{n-1}$	The general power rule
$ax$	$a$	A straight line has constant gradient
$c$ (a constant)	$0$	A constant never changes — its rate of change is zero
$\dfrac{a}{x}=ax^{-1}$	$-ax^{-2}$	Rewrite as a negative power first

y=x^3-3x^2-9x+5 \implies \frac{dy}{dx}=3x^2-6x-9

Exam Tip

Differentiate each term of a sum completely separately — a term like $-9x$ differentiates to the constant $-9$ , and the standalone number $+5$ disappears entirely, since a constant's rate of change is always zero.

03 — Explanation

How do I find a turning point?

Set $\dfrac{dy}{dx}=0$ and solve for $x$ — a curve is momentarily flat (neither rising nor falling) exactly at a turning point.

3x^2-6x-9=0 \implies x^2-2x-3=0 \implies (x-3)(x+1)=0 \implies x=3 \text{ or } x=-1

Exam Tip

To find the matching $y$ -coordinate, substitute back into the original equation $y=f(x)$ — never into $\dfrac{dy}{dx}$ , which is always zero at these points and tells you nothing about the height of the curve.

Substituting into $y=x^3-3x^2-9x+5$ : at $x=3$ , $y=-22$ ; at $x=-1$ , $y=10$ .

04 — Explanation

How do I tell whether a turning point is a maximum or a minimum?

Differentiate a second time to get $\dfrac{d^2y}{dx^2}$ , then substitute the $x$ -value of the turning point into it.

$\dfrac{d^2y}{dx^2}$	Curve shape	Type
$>0$	Concave up ( $\cup$ , like a valley)	Minimum
$<0$	Concave down ( $\cap$ , like a hill)	Maximum

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

At $x=3$ : $6(3)-6=12>0 \implies$ minimum at $(3,-22)$ . At $x=-1$ : $6(-1)-6=-12<0 \implies$ maximum at $(-1,10)$ .

05 — Explanation

What if I can't use the second derivative, or it comes out equal to zero?

Test the sign of $\dfrac{dy}{dx}$ just to the left and just to the right of the turning point instead. If the gradient goes from positive to negative, the curve rises then falls, giving a maximum; if it goes from negative to positive, the curve falls then rises, giving a minimum.

At $x=-1$ for the curve above, testing $x=-1.1$ gives $\dfrac{dy}{dx}=3(-1.1)^2-6(-1.1)-9=1.03>0$ , and testing $x=-0.9$ gives $3(-0.9)^2-6(-0.9)-9=-1.17<0$ — positive then negative confirms a maximum at $x=-1$ , agreeing with the second-derivative test.

Worked Example

A curve has equation $y=2x^3-3x^2-12x+7$ . Find the coordinates of its stationary points and determine the nature of each.

Differentiate: $\dfrac{dy}{dx}=6x^2-6x-12$ .
Set equal to zero and simplify: $6x^2-6x-12=0 \implies x^2-x-2=0 \implies (x-2)(x+1)=0 \implies x=2$ or $x=-1$ .
Substitute into the original equation: at $x=2$ , $y=16-12-24+7=-13$ ; at $x=-1$ , $y=-2-3+12+7=14$ .
Differentiate again: $\dfrac{d^2y}{dx^2}=12x-6$ .
At $x=2$ : $12(2)-6=18>0 \implies$ minimum at $(2,-13)$ . At $x=-1$ : $12(-1)-6=-18<0 \implies$ maximum at $(-1,14)$ .

Exam Tip

A positive second derivative describes a curve that bends upward like a valley — that's a minimum — not a maximum. It's easy to reverse this rule under exam pressure, so it's worth double-checking against the sketch of the curve if time allows, or by using the sign-test method above as a cross-check.