From First Principles

Two points h apart

To find the gradient of a curve y = f(x) at a point, start with a chord. Take the point (x, f(x)) and a second point a distance h further along, (x + h, f(x + h)). The run between them is h and the rise is f(x + h) − f(x), so the gradient of the chord is $\left(f\left(x+h\right)-f\left(x\right)\right)/h$(f(x + h) − f(x)) / h.

On $y=x^2$y = x² at x = 1, with h = 1.4, the two points are (1, 1) and (2.4, 5.76). The rise is 5.76 − 1 = 4.76 and the run is 1.4, so the chord gradient is 4.76 ÷ 1.4 = 3.4.

A chord on $y=x^2$y = x² from x = 1 to x = 2.4, with its run h and its rise marked. Its gradient, the rise divided by the run, is 3.4.

The algebra for $x^2$x²

For $f\left(x\right)=x^2$f(x) = x², the chord gradient is $\left({\left(x+h\right)}^2-x^2\right)/h$((x + h)² − x²) / h. Expand the bracket: ${\left(x+h\right)}^2=x^2+2xh+h^2$(x + h)² = x² + 2xh + h². The $x^2$x² cancels with the $-x^2$− x², so the numerator is $2xh+h^2$2xh + h².

Each term of the numerator has a factor h, and h is not 0 for a chord, so divide through by h: $\left(2xh+h^2\right)/h=2x+h$(2xh + h²) / h = 2x + h. Check it on the chord above: with x = 1 and h = 1.4, 2x + h = 2 + 1.4 = 3.4.

Shrink h

Keep x = 1 and move the second point closer. With h = 0.9 the chord gradient is 2 + 0.9 = 2.9; with h = 0.45 it is 2.45; with h = 0.2 it is 2.2; with h = 0.01 it is 2.01. Each chord lies closer to the tangent at (1, 1) than the one before.

Check one without the formula: with h = 0.2 the second point is (1.2, 1.44), and (1.44 − 1) ÷ 0.2 = 0.44 ÷ 0.2 = 2.2.

Chords from x = 1 with h = 1.4, 0.9, 0.45 and 0.2, each fainter than the last. Their gradients are 3.4, 2.9, 2.45 and 2.2, closing in on the tangent.

At the limit

As h tends to 0, 2x + h tends to 2x. The chords settle onto the tangent, so the gradient of $y=x^2$y = x² at any x is 2x. At x = 1 it is 2, and at x = 3 it is 6: the tangent at (3, 9) has gradient 6.

This limit is called the derivative of f, and it is written f'(x) or $\frac{dy}{dx}$dy/dx: f'(x) = the limit of $\left(f\left(x+h\right)-f\left(x\right)\right)/h$(f(x + h) − f(x)) / h as h tends to 0. Working it out from the chord, as here, is called differentiating from first principles. For $f\left(x\right)=x^2$f(x) = x², f'(x) = 2x.

At the limit only the tangent at (1, 1) is left. Its gradient is 2x with x = 1, which is 2.

At x = 2

At x = 2 the chord gradient is 2x + h = 4 + h. Slide h down to 0 and the chord becomes the tangent at (2, 4), with gradient 2 × 2 = 4.

Drag the upper point down the curve toward (2, 4). The gap h shrinks, the dashed chord turns toward the tangent, and at h = 0 the gradient is f'(2) = 4.

Two more functions

For the straight line f(x) = 3x + 1, the rise is 3(x + h) + 1 − (3x + 1) = 3h, and $3h/h=3$3h / h = 3. No h is left to shrink, so f'(x) = 3 at every x: a straight line has the same gradient everywhere.

For $f\left(x\right)=x^2-4x$f(x) = x² − 4x, the rise is ${\left(x+h\right)}^2-4\left(x+h\right)-\left(x^2-4x\right)=2xh+h^2-4h$(x + h)² − 4(x + h) − (x² − 4x) = 2xh + h² − 4h. Dividing by h gives 2x + h − 4, which tends to 2x − 4. At x = 2 the gradient is 0: this is the bottom of the curve, where the tangent is level. Check with h = 0.01: f(2.01) = 4.0401 − 8.04 = −3.9999 and f(2) = −4, so the chord gradient is 0.0001 ÷ 0.01 = 0.01, close to 0.

Why h goes to 0 last

Putting h = 0 into $\left({\left(x+h\right)}^2-x^2\right)/h$((x + h)² − x²) / h at the start gives 0 ÷ 0, which has no value: with h = 0 there is only one point, and no chord. The h in the denominator has to be divided out first, while h is still not 0. Only then can h be let tend to 0.

The usual mistakes

Giving the height for the gradient. At x = 3 on $y=x^2$y = x² the height is 9, but the gradient is 2x = 6.

Giving x for the gradient of $x^2$x². The 2 in 2x comes from the 2xh term of the expansion, and it cannot be dropped: at x = 3 the gradient is 6, not 3.

Expanding ${\left(x+h\right)}^2$(x + h)² as $x^2+h^2$x² + h². The middle term 2xh is the one the derivative comes from.

Putting h = 0 before dividing through by h, which gives 0 ÷ 0.

A heated cube

In the application below, a metal cube has volume $V=x^3$V = x³ when its edge is x. First principles on $x^3$x³ gives the rate at which the volume grows with the edge, and that rate estimates the growth for a small change.