Free Pearson Edexcel International A-level Mathematics · P1

Differentiation, tangents and normals

What you’ll learn

  • Differentiate sums of rational powers.
  • Interpret a derivative as gradient and rate of change.
  • Find tangent and normal equations.

Understand the idea

A derivative gives the instantaneous rate of change and the gradient of the tangent. It arises as the limit of a secant gradient as the two points approach each other. Differentiate a power by multiplying by its exponent and reducing the exponent by one; constants differentiate to zero.

Expand products and rewrite roots or quotients as powers before using the power rule in P1. Differentiate again to obtain the second derivative. To find a tangent, calculate both the point on the curve and the gradient there, then use the straight-line equation. The normal is perpendicular to the tangent.

Key formulae

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h

ddx(xn)=nxn−1\frac{d}{dx}(x^n)=nx^{n-1}

y−y0=f′(x0)(x−x0)y-y_0=f'(x_0)(x-x_0)

mnormal=−1f′(x0)(f′(x0)≠0)m_{\mathrm{normal}}=-\frac1{f'(x_0)}\quad(f'(x_0)\ne0)

A tangent matches the local rate of change

Conditions. Curve y = x² and its tangent y = 2x − 1 at (1,1). Both are plotted on −1 ≤ x ≤ 3.

x / dimensionless; y / dimensionless.

At x = 1 both curves pass through y = 1 and have gradient 2. The tangent is not the same as the whole parabola; away from the contact point its height differs by (x − 1)².

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

Find the tangent to y=x3−2xy=x^3-2x at x=2x=2.

  1. The point has y=8−4=4y=8-4=4, so it is (2,4)(2,4).
  2. Differentiate: dy/dx=3x2−2dy/dx=3x^2-2. At x=2x=2 the gradient is 1010.
  3. Substitute into point-gradient form: y−4=10(x−2)y-4=10(x-2).

Answer: y=10x−16y=10x-16.

Open the interactive lesson to try original questions and recall. Build a revision plan.