A-level · Beta

Differentiation: from secant to tangent

Move points on a polynomial and connect shrinking intervals to its derivative.

Drag P along the curve; drag Q sideways to change the interval.

f(x)=x^{2}

Tab to a handle, then use the arrow keys. Use the sliders for precise values.

The red secant joins P and Q. The green dashed line is the tangent at P.

As h approaches zero, the secant gradient approaches the derivative. The interval stays positive; displayed decimals are rounded to 4 places.

Predict, test and explain

Step 1 of 3

Predict

For f(x) = x³, what is the tangent gradient at x = 2?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Predict

Try it in the simulation. Change one thing at a time.

Explain

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

For f(x) = x³, what is the tangent gradient at x = 2?

$f(x)=x^3,\quad f'(2)=?$

Hint

A secant uses two distinct points. A tangent uses the limiting gradient.

Reveal answer

The derivative is 3x², so at x = 2 it is 3 × 4 = 12.

$12$

Hint 1

A secant uses two distinct points. A tangent uses the limiting gradient.

Hint 2

Differentiate axⁿ as naxⁿ⁻¹.

Hint 3

Evaluate the derivative at the point; the function value gives height, not gradient.

Worked example

This example uses fixed values, separate from the diagram controls.

Find the tangent to y = x² at x = 2.

Answer: y = 4x − 4.

Connect this idea

Watch out: The curve height f(2) = 8 is not the gradient f′(2) = 12.

Specification and learning route

AQA G1, G2, G3 · Edexcel Pure 7.1, Pure 7.2, Pure 7.3

AS core; A-level extension. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Gradient; Algebra and indices; Functions and graphs.