The Derivative As a Limit

The gradient of a chord

A chord is a straight line through two points of a curve. Its gradient is an ordinary rise over run. On the curve $y=x^2$y = x², take the points (2, 4) and (3, 9). The run from one to the other is 3 − 2 = 1 and the rise is 9 − 4 = 5, so the chord has gradient 5 ÷ 1 = 5.

A chord gradient is the average steepness of the curve between its two ends. The curve $y=x^2$y = x² gets steeper as x grows, so between x = 2 and x = 3 it is less steep than 5 at the left end and steeper than 5 at the right end.

The chord of $y=x^2$y = x² through (2, 4) and (3, 9). It rises 5 over a run of 1, so its gradient is 5.

Slide one end closer

Keep the end at (2, 4) and move the other end toward it. With the second point at x = 2.4, the height there is $2.4^2=5.76$2.4² = 5.76. The rise is 5.76 − 4 = 1.76 and the run is 0.4, so the chord gradient is 1.76 ÷ 0.4 = 4.4.

Call the run h, so the second point is at x = 2 + h. The chord gradient is $\left({\left(2+h\right)}^2-4\right)/h$((2 + h)² − 4) / h. With h = 1 it is 5, with h = 0.4 it is 4.4, and with h = 0.1 it is (4.41 − 4) ÷ 0.1 = 4.1. As the second point slides toward (2, 4), the chord swings round and lies closer and closer to the tangent at (2, 4).

The second point may also sit to the left. With h = −0.1 it is at x = 1.9, where the height is 3.61; the rise is 3.61 − 4 = −0.39 and the run is −0.1, so the gradient is 3.9.

The chord from (2, 4) to (2.4, 5.76). Its run is 0.4 and its rise is 1.76, so its gradient is 4.4: less steep than the chord to (3, 9), and closer to the tangent at (2, 4).

Chord gradients of $y=x^2$y = x² for chords of width h with one end at x = 2. Each row is one width h, from 1 down to 0.001; the first column has the chord to the right of 2 and the second the chord to the left. From the right the gradients fall toward 4, and from the left they rise toward 4.

Where the swinging settles

The table suggests 4, and algebra shows it. Expand the rise: ${\left(2+h\right)}^2-4=4+4h+h^2-4=4h+h^2$(2 + h)² − 4 = 4 + 4h + h² − 4 = 4h + h². For a chord, h is not 0, so the h in the denominator can be divided out: $\left(4h+h^2\right)/h=4+h$(4h + h²) / h = 4 + h.

So every chord through (2, 4) has gradient exactly 4 + h. Check it on the table: with h = 0.01 that is 4.01, and with h = −0.001 it is 3.999. Now let h tend to 0. The expression 4 + h has no h in a denominator, so its limit is found by substituting: 4 + 0 = 4.

That limit is the gradient of the tangent at (2, 4), and it is called the derivative of $y=x^2$y = x² at x = 2. For any function f, the derivative at a point a is the limit of the chord gradient $\left(f\left(a+h\right)-f\left(a\right)\right)/h$(f(a + h) − f(a)) / h as h tends to 0, from both sides.

Chords of $y=x^2$y = x² from x = 2 with widths 1, 0.4 and 0.1, each fainter than the last. Their gradients are 5, 4.4 and 4.1, and they close on the tangent at (2, 4), whose gradient is 4.

The chord of $y=x^2$y = x² from (2, 4) to $\left(2+h,{\left(2+h\right)}^2\right)$(2 + h, (2 + h)²). At h = 2 the second point is (4, 16) and the gradient is 12 ÷ 2 = 6, which is 4 + h. Drag the upper point down the curve: the gradient falls by exactly as much as h does, and at h = 0 the chord has become the tangent, with gradient 4.

A limit, not an approximation

A chord with h = 0.001 has gradient 4.001. That is an approximation to the derivative, and a good one, but it is not the derivative. Every chord gradient is 4 + h with h not 0, so no chord has gradient exactly 4. The derivative is 4 exactly, because 4 is the one number that 4 + h approaches as h tends to 0, from either side.

Nor can h be put equal to 0 at the start. In $\left({\left(2+h\right)}^2-4\right)/h$((2 + h)² − 4) / h, h = 0 gives $\left(4-4\right)÷0=\frac{0}{0}$(4 − 4) ÷ 0 = 0/0, which is not a number: with h = 0 there is only one point, and one point makes no chord. The h has to be divided out while it is still not 0, and only then let tend to 0.

At another point

At (3, 9) the chord to x = 3 + h rises ${\left(3+h\right)}^2-9=6h+h^2$(3 + h)² − 9 = 6h + h², so its gradient is $\left(6h+h^2\right)/h=6+h$(6h + h²) / h = 6 + h. With h = 0.1 that is (9.61 − 9) ÷ 0.1 = 6.1. As h tends to 0, 6 + h tends to 6, so the derivative of $y=x^2$y = x² at x = 3 is 6.

The same working at any x gives $\left(2xh+h^2\right)/h=2x+h$(2xh + h²) / h = 2x + h, which tends to 2x. The derivative at x = 2 is 4 and at x = 3 is 6, and both are 2x.

The usual mistakes

Giving a chord gradient for the derivative. The chord from x = 2 to x = 3 has gradient 5, but the derivative at x = 2 is 4. A chord uses two points and the derivative uses one.

Giving the height for the gradient. At x = 3 the curve $y=x^2$y = x² has height 9, but the chord from 3 to 4 has gradient (16 − 9) ÷ 1 = 7 and the derivative at 3 is 6. A gradient comes from a rise between two points.

Calling the derivative "about 4" because the chord gradients get close to 4. The limit is exactly 4; 4.001 is one chord's gradient, not the limit.

Putting h = 0 before dividing by h, which gives $\frac{0}{0}$0/0 and no answer at all.