A car passes kilometre 120 at 2:00 p.m. and kilometre 210 at 3:00 p.m. Its average speed was

$$\frac{\Delta s}{\Delta t}=\frac{210-120}{15-14}=90\ \text{km/h}.$$

The Greek letter $\Delta$, capital delta, is the Greek $D$, for difference. $\Delta s$ is not “delta times $s$”: it is a single sign meaning “how much $s$ has changed”. It is a calculation with real numbers, one you can do with a ruler and a clock.

But the speedometer does not show the average speed over an hour. It shows the speed now. And “now” has no duration: $\Delta t=0$, and dividing by zero makes no sense. Calculus resolves this impasse with a passage to the limit, and its notation follows the passage without changing shape.

The fraction that shrinks

Take $y=x^2$ and a point $x=a$. Moving $\Delta x$ to the right, $y$ changes by

$$\Delta y=(a+\Delta x)^2-a^2=2a\,\Delta x+(\Delta x)^2,$$

and the average rate of change is

$$\frac{\Delta y}{\Delta x}=2a+\Delta x.$$

Now shrink $\Delta x$. The rate approaches $2a$, and the remainder, $\Delta x$, vanishes. What is left gets Leibniz’s name:

$$\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=2a.$$

Notice what the notation does. The fraction $\frac{\Delta y}{\Delta x}$ becomes the fraction $\frac{dy}{dx}$: the shape stays, and only the Greek letter turns Latin. The $d$ is a $\Delta$ that has shrunk. This visual continuity is an enormous asset: anyone who knows how to handle the fraction of finite differences already has an intuition about how to handle the derivative. And it is also the source of the mistakes: the derivative is not a fraction of two numbers, and the next text shows where the resemblance breaks down.

Δy is not dy

In modern notation, $\Delta y$ and $dy$ are different things, and it is worth keeping them apart. For a step $\Delta x$:

  • $\Delta y=f(x+\Delta x)-f(x)$ is the true change in the function;
  • $dy=f'(x)\,\Delta x$ is the change along the tangent line, the linear approximation.

For $y=x^2$ at $x=3$ with $\Delta x=0.1$: $\Delta y=3.1^2-3^2=0.61$, while $dy=6\cdot0.1=0.6$. The difference, $0.01=(\Delta x)^2$, is small compared with $\Delta x$ and becomes ever smaller as the step shrinks. That is what justifies using $dy$ instead of $\Delta y$ in approximate calculations.

This approximation is one of the most used tools in science. What is $\sqrt{4.1}$? With $f(x)=\sqrt x$, $f'(x)=\frac1{2\sqrt x}$, and at $x=4$:

$$\sqrt{4.1}\approx\sqrt4+\frac{1}{2\sqrt4}\cdot0.1=2+0.025=2.025.$$

The true value is $2.02485\ldots$ The error is in the fifth decimal place.

Exercise: Δ for y = x³

Calculate $\frac{\Delta y}{\Delta x}$ for $y=x^3$ at $a=1$ with $\Delta x=0.1$, and then the limit.

$\Delta y=1.1^3-1=1.331-1=0.331$, so $\frac{\Delta y}{\Delta x}=3.31$. In general, $\frac{(1+\Delta x)^3-1}{\Delta x}=3+3\Delta x+(\Delta x)^2$, which tends to $3$. The derivative of $x^3$ at $x=1$ is $3\cdot1^2=3$.

Exercise: use dy instead of Δy

Estimate $1.02^5$ with the differential.

With $f(x)=x^5$, $f'(x)=5x^4$, and at $x=1$ with $\Delta x=0.02$: $dy=5\cdot0.02=0.1$, so $1.02^5\approx1.1$. The exact value is $1.10408\ldots$ The error comes from the terms in $(\Delta x)^2$ and higher powers, which the differential ignores.

Chapter index · ← Four ways to write the derivative · Is dy/dx a fraction? →