Product and quotient rules

The wrong rule, made obvious

The sum rule was so obliging — derivative of a sum is the sum of derivatives — that you might expect the same for products. Surely the derivative of a product is the product of the derivatives?

It is not. The cleanest disproof is one line. Take $f(x) = x \cdot x$ . The wrong rule predicts $f'(x) = 1 \cdot 1 = 1$ . The right answer (you already know) is $f'(x) = 2x$ .

So whatever the rule for products is, it has to recover $2x$ from $x \cdot x$ . We are about to derive it geometrically, and the answer will write itself.

Two strips plus a vanishing corner

A product $u \cdot v$ is the area of a rectangle with sides $u$ and $v$ . Drag the parameter $t$ below. Both $u(t)$ and $v(t)$ grow a little. The new area shows up as three pieces, not two.

A top strip of width $u$ and height $dv$ . Area: $u \cdot dv$ .
A right strip of height $v$ and width $du$ . Area: $v \cdot du$ .
A tiny corner square of size $du \cdot dv$ .

Now shrink $dt$ . The two strips shrink linearly with $dt$ . The corner shrinks quadratically — it has both $du$ and $dv$ in it, each going to zero. In the limit, the corner vanishes and the two strips survive.

The product rule

Divide the change in area by $dt$ and send $dt \to 0$ :

\boxed{\;(uv)' \;=\; u' \, v \;+\; u \, v'\;}

Read it as: derivative of the first times the second, plus the first times the derivative of the second. The “plus” matters. The wrong rule misses the second term entirely.

Sanity-check it on $f(x) = x \cdot x$ with $u = v = x$ , so $u' = v' = 1$ :

(x \cdot x)' \;=\; 1 \cdot x + x \cdot 1 \;=\; 2x. \quad\checkmark

And on a real example, $f(x) = x^2 \sin x$ :

\frac{d}{dx}\bigl(x^2 \sin x\bigr) \;=\; 2x \cdot \sin x \;+\; x^2 \cdot \cos x.

Both terms have to be there.

A product, at a point

Differentiate $f(x) = x^2 \sin x$ using the product rule.

What is $f'(0)$ ?

Quotients

Division is just multiplication by a reciprocal. One quick-and-dirty option is to write $f/g = f \cdot g^{-1}$ and use the product rule with the power rule (Karpathy’s micrograd does exactly this and never needs a dedicated division op). But there is also a direct rule worth knowing:

\left(\frac{f}{g}\right)' \;=\; \frac{f' \, g \;-\; f \, g'}{g^2}.

The mnemonic is centuries old: low d-high minus high d-low, square the bottom and away we go. The order in the numerator matters — flip $f'g$ and $fg'$ and you get the negative of the right answer.

A quotient, at a point

Let $f(x) = \dfrac{x^2 + 1}{x - 3}$ .

Compute $f'(0)$ to three decimals.

Two more rules in the bag

You now have:

(uv)' = u' v + u v' \qquad \left(\frac{u}{v}\right)' = \frac{u' v - u v'}{v^2}.

These let you differentiate any product or ratio of the elementary functions from the previous lesson. What you cannot yet do is differentiate a composition — a function inside another, like $\sin(x^2)$ or $e^{-x^2}$ or the sigmoid $1/(1 + e^{-x})$ that every neural network in the world is built on.

That is the chain rule, and it is the load-bearing piece of this entire module.