Differentiating From First Principles

The difference quotient

Every derivation from first principles starts in the same place. Take a point x on the curve y = f(x) and a second point a step h further along, at x + h. The rise between them is f(x + h) − f(x), and the run is h.

Dividing the rise by the run gives the gradient of the chord: $\left(f\left(x + h\right) - f\left(x\right)\right) / h$. This is called the difference quotient. The derivative is its limit as h tends to 0: $f'\left(x\right) = \lim \left(f\left(x + h\right) - f\left(x\right)\right) / h$ as $h \to 0$.

The method has four moves, always in this order: write the rise out in full, simplify it, divide by h, and only then let h tend to 0.

On $y = x^{2}$, the step from x = 1 to x = 2.5 has run h = 1.5. The rise is f(2.5) − f(1) = 6.25 − 1 = 5.25, so the difference quotient is 5.25 ÷ 1.5 = 3.5, the gradient of the chord.

Write the difference out in full

For $f\left(x\right) = x^{2}$, the rise is $\left(x + h\right)^{2} - x^{2}$. Expand the bracket: $\left(x + h\right)^{2} = x^{2} + 2xh + h^{2}$, and the $x^{2}$ cancels with the $- x^{2}$, leaving $2xh + h^{2}$.

Every term that is left has a factor h. A chord has h not equal to 0, so dividing by h is allowed: $\left(2xh + h^{2}\right) / h = 2x + h$. Check it on the chord above: with x = 1 and h = 1.5, 2x + h = 2 + 1.5 = 3.5, the gradient found from the two points.

Only now let h vanish

The chord gradient is now 2x + h, with no h in a denominator. As h tends to 0 it tends to 2x, so f′(x) = 2x. At x = 1 the chord gradients 3.5, 3 and 2.5, for h = 1.5, 1 and 0.5, close in on 2.

The order matters. Putting h = 0 into $\left(\left(x + h\right)^{2} - x^{2}\right) / h$ at the start gives $\left(x^{2} - x^{2}\right) / 0 = \frac{0}{0}$, which is not a number. Dividing first removes the h that makes the denominator 0, and after that, putting h = 0 into 2x + h is safe.

Chords of $y = x^{2}$ from x = 1 with h = 1.5, 1 and 0.5, each fainter than the last. Their gradients are 2x + h = 3.5, 3 and 2.5, and they close on the tangent at (1, 1), whose gradient is 2.

A cube

For $f\left(x\right) = x^{3}$, the rise is $\left(x + h\right)^{3} - x^{3}$. Expand: $\left(x + h\right)^{3} = x^{3} + 3x^{2}h + 3xh^{2} + h^{3}$, so the rise is $3x^{2}h + 3xh^{2} + h^{3}$ once the $x^{3}$ cancels.

Divide by h: $3x^{2} + 3xh + h^{2}$. Every term after the first still carries an h, so as h tends to 0 they tend to 0, and $f'\left(x\right) = 3x^{2}$.

Check it at x = 2, where $3x^{2} = 12$. With h = 0.1 the quotient is $\left(2.1^{3} - 8\right) / 0.1 = \left(9.261 - 8\right) / 0.1 = 12.61$, which is $12 + 3 \times 2 \times 0.1 + 0.1^{2} = 12 + 0.6 + 0.01$. The table shows both sides closing in on 12.

Chord gradients of $y = x^{3}$ for chords of width h with one end at x = 2. Each row is one width h; the first column has the chord to the right of 2 and the second the chord to the left. They close in on 12 from above and from below, as $3x^{2} + 3xh + h^{2}$ says they must.

A line, and a curve lifted by 1

For f(x) = 3x, the rise is 3(x + h) − 3x = 3h, and 3h / h = 3. No h is left, so there is nothing to wait for: the derivative is 3 at every x. It is not 0, because 3x climbs 3 for every 1 across.

For $f\left(x\right) = x^{2} + 1$, the rise is $\left(x + h\right)^{2} + 1 - \left(x^{2} + 1\right)$. The two 1s cancel, leaving $\left(x + h\right)^{2} - x^{2} = 2xh + h^{2}$, exactly as for $x^{2}$. So the derivative is 2x again: the + 1 lifts the curve and does not change its gradient anywhere.

A reciprocal

For $f\left(x\right) = \frac{1}{x}$, the rise is $\frac{1}{{x + h}} - \frac{1}{x}$. Put it over the common denominator x(x + h): $\left(x - \left(x + h\right)\right) / \left(x\left(x + h\right)\right) = -h / \left(x\left(x + h\right)\right)$.

Divide by h: $-1 / \left(x\left(x + h\right)\right)$. As h tends to 0, x + h tends to x, so $f'\left(x\right) = - \frac{1}{x^{2}}$. The minus sign says the curve falls everywhere, for x positive or negative.

Check it at x = 2, where $- \frac{1}{x^{2}} = -0.25$. With h = 0.1 the quotient is $\left(\frac{1}{2.1} - \frac{1}{2}\right) \div 0.1 = -0.238095$; with h = 0.01 it is −0.248756; with h = 0.001 it is −0.249875.

The curve $y = \frac{1}{x}$ and its tangent at (2, 0.5), the line y = −0.25x + 1. Its gradient is $- \frac{1}{x^{2}}$ at x = 2, which is −0.25.

A square root

For $f\left(x\right) = \sqrt{x}$, the quotient is $\left(\sqrt{x + h} - \sqrt{x}\right) / h$, and no expansion clears the h. Multiply the numerator and the denominator by $\sqrt{x + h} + \sqrt{x}$. The numerator becomes a difference of two squares: $\left(\sqrt{x + h} - \sqrt{x}\right)\left(\sqrt{x + h} + \sqrt{x}\right) = \left(x + h\right) - x = h$.

So the quotient is $h / \left(h\left(\sqrt{x + h} + \sqrt{x}\right)\right) = 1 / \left(\sqrt{x + h} + \sqrt{x}\right)$, and as h tends to 0 this tends to $1 / \left(2\sqrt{x}\right)$. At x = 4 that is $\frac{1}{4} = 0.25$. With h = 0.1 the quotient is 0.248457, with h = 0.01 it is 0.249844, and with h = 0.001 it is 0.249984.

The usual mistakes

Putting h = 0 before dividing by h. That gives $\frac{0}{0}$, which decides nothing; the h in the denominator must be divided out first.

Expanding $\left(x + h\right)^{2}$ as $x^{2} + h^{2}$. The middle term 2xh is lost, and it is the term the derivative comes from.

Writing f(x) + h for f(x + h). For $x^{2}$ that gives a rise of $\left(x^{2} + h\right) - x^{2} = h$ and a quotient of 1, which is wrong: f(x + h) means x + h put in place of x.

Multiplying by the power without lowering it. For $x^{2}$ the surviving term is 2xh, one power of x down, so the answer is 2x and not $2x^{2}$.

Leaving h in the answer. 2x + h is the gradient of a chord; the derivative is its limit, 2x.

A reservoir

In the application below, the depth of a reservoir in a dry spell is $120 - 0.5x^{2}$ centimeters on day x. The rise over a step h is worked out in full, divided by h, and only then is h let tend to 0, giving the rate the depth falls. On day 6 the rain starts, and the quotients from the two sides of day 6 tend to two different numbers.