Calculus I · The Slope of a Single Point

The derivative

Here is the definition the entire back half of this book rests on.

The derivative of $f$ at the point $a$, written $f'(a)$, is $$f'(a) = \lim_{h\to 0}\frac{f(a+h) - f(a)}{h},$$ when this limit exists. It is the instantaneous rate of change of $f$ at $a$, and the slope of the tangent line to the graph at $a$.

That’s it. The derivative is the limit of the average rate as the interval shrinks to nothing. Three names for one number: instantaneous rate of change, slope of the tangent line, the derivative. Now let’s actually compute some — and notice that every single computation is the same dance: write the difference quotient, simplify until the $h$ in the denominator cancels, then let $h \to 0$.

Computing derivatives by hand

The slope of $x^2$. Let $f(x) = x^2$. Build the difference quotient at a general point $a$ and simplify:

$$\frac{(a+h)^2 - a^2}{h} = \frac{a^2 + 2ah + h^2 - a^2}{h} = \frac{2ah + h^2}{h} = \frac{h(2a + h)}{h} = 2a + h.$$

The $h$ canceled — the $\tfrac00$ dissolved, exactly as in Chapter 2. Now let $h \to 0$:

$$f'(a) = \lim_{h\to 0}(2a + h) = 2a.$$

So the slope of $y = x^2$ at any point $a$ is $2a$. At $x = 3$ the slope is $6$; at $x = 0$ it’s $0$ (the bottom of the parabola is flat); at $x = -1$ it’s $-2$ (sloping down). You computed a derivative — and if you did the Chapter 2 “Looking ahead” problems, you’ve seen this exact calculation before; now it has a name.

The slope of $x^3$. Same dance, one more term to expand:

$$\frac{(a+h)^3 - a^3}{h} = \frac{3a^2 h + 3a h^2 + h^3}{h} = 3a^2 + 3ah + h^2 \;\xrightarrow{\,h\to0\,}\; 3a^2.$$

So the derivative of $x^3$ is $3x^2$.

The slope of $\tfrac1x$. Now a fraction. Let $f(x) = \tfrac1x$. The difference quotient is a “fraction of fractions” — combine the top over a common denominator first:

$$\frac{\frac{1}{a+h} - \frac{1}{a}}{h} = \frac{\frac{a - (a+h)}{a(a+h)}}{h} = \frac{-h}{h\,a(a+h)} = \frac{-1}{a(a+h)} \;\xrightarrow{\,h\to0\,}\; \frac{-1}{a^2}.$$

So the derivative of $\tfrac1x$ is $-\tfrac{1}{x^2}$ — always negative, because $\tfrac1x$ always slopes downward.

The slope of $\sqrt{x}$. A square root: rationalize, the trick from Chapter 2. Let $f(x) = \sqrt{x}$ and multiply top and bottom by the conjugate $\sqrt{a+h} + \sqrt{a}$:

$$\frac{\sqrt{a+h} - \sqrt{a}}{h} = \frac{(a+h) - a}{h\left(\sqrt{a+h} + \sqrt{a}\right)} = \frac{1}{\sqrt{a+h} + \sqrt{a}} \;\xrightarrow{\,h\to0\,}\; \frac{1}{2\sqrt{a}}.$$

So the derivative of $\sqrt{x}$ is $\tfrac{1}{2\sqrt{x}}$.

The two simplest of all. A constant function $f(x) = c$ never changes, so its difference quotient is $\tfrac{c - c}{h} = 0$ — derivative $0$, as it must be (a flat line has slope $0$). A line $f(x) = mx + b$ has difference quotient $\tfrac{[m(a+h)+b] - [ma+b]}{h} = \tfrac{mh}{h} = m$ — derivative $m$, the slope it always had. A line is its own slope; that’s the whole point of a line.

Toolkit: Computing a derivative from the definition

1. Write the difference quotient $\dfrac{f(a+h) - f(a)}{h}$.
2. Simplify until the $h$ in the denominator cancels — expand and combine ($x^2, x^3$), find a common denominator (fractions), or rationalize (roots).
3. Let $h \to 0$ in what’s left.

Every derivative in this chapter is these three steps. The algebra changes; the dance never does.

All 12 chapters of Calculus I · Foundations · All courses