12.4PureCore

Differentiating Quadratics

A quadratic differentiates term by term using the power rule. The result is a linear gradient function, which lets you find the gradient anywhere on the parabola — and where it is flat.

20 min Video by Zeeshan Zamurred Differentiation

Edexcel AS Level Maths: 12.4 Differentiating QuadraticsWatch the full walkthrough before the notes below.

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What you'll be able to do

Differentiate a quadratic term by term
Find the gradient at a specific point
Find where the gradient is zero
Interpret the gradient on a parabola

Term by term

Differentiate each term of $a x^{2} + b x + c$ separately. The $x^{2}$ becomes $2 a x$ , the $b x$ becomes $b$ , and the constant disappears.

y = a x^{2} + b x + c \Rightarrow \frac{d y}{d x} = 2 a x + b

A linear gradient function.

3 x^{2} \to 6 x

;

- 4 x \to - 4

;

7 \to 0

Answer

\frac{d y}{d x} = 6 x - 4

Gradient at a point

Substitute the $x$ -value into the gradient function to get the gradient there.

\frac{d y}{d x} = 2 x - 2

2At

x = 3

2 (3) - 2 = 4

Answergradient

= 4

Where the gradient is zero

Setting the gradient function to zero finds the $x$ where the parabola is flat — its turning point. This previews stationary points.

\frac{d y}{d x} = 2 x - 6 = 0

x = 3

Answer

x = 3

(the vertex)

Tip — Zero gradient ⟶ the turning point. This matches the line of symmetry you found by completing the square.

Formula recap

y = a x^{2} + b x + c \Rightarrow \frac{d y}{d x} = 2 a x + b

Differentiate a quadratic.

\frac{d y}{d x} = 0 \Rightarrow turning point

Flat gradient at the vertex.

Common mistakes to avoid

Keeping the constant term in the derivative.

The constant differentiates to 0 and disappears.

Forgetting the coefficient when differentiating ax².

ax² → 2ax (the 2 and the a both matter).

Key takeaways

ax² + bx + c differentiates to 2ax + b.
Substitute an x-value to get the gradient at a point.
Setting dy/dx = 0 locates the turning point.

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