Math · Calculus I · Concept

The limit definition of the derivative

The derivative $f^{'} (a)$ is the slope of the tangent line to $y = f (x)$ at $x = a$ , and the instantaneous rate of change of $f$ there. It is the limit of the secant slopes $(f (a + h) - f (a)) / h$ as $h$ approaches $0$ , when that limit exists.

A concrete case

How fast is the stone falling?

A stone dropped from a cliff falls about $5 t^{2}$ meters in $t$ seconds. After $2 seconds$ it has fallen $20$ meters, and after $3 seconds$ $45$ meters. Over that second it averaged $(45 - 20) / 1 = 25$ meters per second.

That is an average over a whole second, and the stone speeds up the entire time. To find its speed at the instant $t = 2$ , shrink the interval. From $2$ to $2.1 seconds$ it falls $22.05 - 20 = 2.05$ meters, an average of $2.05 / 0.1 = 20.5$ meters per second. From $2$ to $2.01 seconds$ the average is $20.05$ , and from $2$ to $2.001$ it is $20.005$ .

The averages close in on $20$ . The speed at the instant $t = 2$ is $20$ meters per second, the number a speedometer would show. It is a limit of averages, and that limit is the derivative.

\frac{5 {(2.1)}^{2} - 5 {(2)}^{2}}{0.1} = \frac{2.05}{0.1} = 20.5

Average speed of the stone from $t = 2$ to $t$ $= 2 + h$
$h$ $(s)$	Distance fallen $(m)$	Average speed $(m / s)$
$1$	$25$	$25$
$0.1$	$2.05$	$20.5$
$0.01$	$0.2005$	$20.05$
$0.001$	$0.020005$	$20.005$

The rule

Secant slopes and the derivative

Pick two inputs on the graph of $f$ , $a$ and $a + h$ . The line through $(a, f (a))$ and $(a + h, f (a + h))$ is a secant line. Its slope is the change in output over the change in input, $(f (a + h) - f (a)) / h$ . This is the difference quotient, and it is the average rate of change of $f$ between the two inputs.

The derivative of $f$ at $a$ , written $f^{'} (a)$ and read “ $f$ prime of $a$ ,” is the limit of that slope as $h$ approaches $0$ . When the limit exists, $f$ is differentiable at $a$ . The line through $(a, f (a))$ with slope $f^{'} (a)$ is the tangent line.

$f^{'} (a)$ has the units of output per unit of input. For the stone, with $s (t) = 5 t^{2}$ , $s^{'} (2) = 20$ meters per second.

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h}

The tangent line and the derivative function

The tangent line at $x = a$ passes through $(a, f (a))$ with slope $f^{'} (a)$ . Point-slope form writes it directly.

Leave $a$ as a letter and the same limit gives the slope at every input at once. The result is a new function, the derivative $f^{'} (x)$ . For $f (x) = x^{2}$ , the limit gives $f^{'} (x) = 2 x$ , so the slope is $4$ at $x = 2$ and $- 2$ at $x = - 1$ . The derivative is also written $d y / d x$ or $d / d x f (x)$ , and $d / d x$ is read “the derivative with respect to $x$ .”

y = f (a) + f^{'} (a) (x - a)

Why it works

Why the limit gives the slope at a point

A secant through two points of a curve has a slope you can compute. A tangent touches at one point, and one point gives no slope by itself. The limit bridges the gap. As $h$ shrinks, the second point slides along the curve toward $(a, f (a))$ , and the secant pivots about that point. If its slope settles on one number, the line settles on one line, and that line is the tangent.

The algebra shows the settling exactly. For the stone, $5 {(2 + h)}^{2} - 20$ $= 20 h + 5 h^{2}$ , and dividing by $h$ gives $20 + 5 h$ for every $h \neq 0$ . The $5 h$ is the extra speed the stone gains during the interval. It shrinks with $h$ , and only $20$ is left.

You cannot get there by setting $h = 0$ at the start, because then the quotient is $0 / 0$ . Simplify while $h \neq 0$ , cancel the $h$ , and then let $h$ approach $0$ .

\frac{5 {(2 + h)}^{2} - 20}{h} = \frac{20 h + 5 h^{2}}{h} = 20 + 5 h

\begin{aligned} \frac{5 {(2 + h)}^{2} - 20}{h} & = \frac{20 h + 5 h^{2}}{h} \\ = 20 + 5 h \end{aligned}

The method

The method in five steps

The table finds the stone’s speed at $t = 2$ from $s (t) = 5 t^{2}$ . The same steps work for any function the algebra can handle.

The cancellation in step 3 is the whole point. If no factor of $h$ cancels, the limit still has the form $0 / 0$ , and the rewrite is not finished.

The limit definition, step by step
Step	What to do	For $s (t)$ $= 5 t^{2}$ at $t = 2$
1	Write $f (a + h)$ and $f (a)$	$s (2 + h)$ $= 5 {(2 + h)}^{2}$ , $s (2)$ $= 20$
2	Subtract and simplify the numerator	$5 (4 + 4 h + h^{2}) - 20$ $= 20 h + 5 h^{2}$
3	Divide by $h$ and cancel, for $h \neq 0$	$20 + 5 h$
4	Let $h$ approach $0$	$20$
5	State the slope with its units	$s^{'} (2)$ $= 20$ meters per second

Opening the $0 / 0$

Step 2 depends on the function. A polynomial is expanded, as above. A fraction such as $1 / x$ needs a common denominator on top, and a square root needs the conjugate. The goal is always the same, a factor of $h$ in the numerator that cancels the $h$ underneath.

How the numerator opens
$f (x)$	Rewrite	Difference quotient at $a$ , for $h \neq 0$
$x^{2}$	expand ${(a + h)}^{2}$	$2 a + h$
$1 / x$	common denominator $a (a + h)$	$- 1 / (a (a + h))$
$\sqrt{x}$	multiply by $\sqrt{a + h} + \sqrt{a}$	$1 / (\sqrt{a + h} + \sqrt{a})$

Where it breaks

The limit can fail, and then $f$ has no derivative at $a$ . At a corner the two sides disagree. For $∣ x ∣$ at $0$ , the quotient $∣ h ∣ / h$ is $1$ for every positive $h$ and $- 1$ for every negative $h$ , and no single slope exists.

At a vertical tangent the quotient grows without bound. For the cube root $\sqrt[3]{x}$ at $0$ , the quotient is $\sqrt[3]{h} / h$ , which is larger than $100$ once $h$ is within $0.001$ of $0$ . A jump or a hole rules out a derivative too, because a function must be continuous at $a$ to be differentiable there.

The reverse is false. $∣ x ∣$ is continuous at $0$ and still has no derivative there. Continuity is necessary, not sufficient.

A single secant slope is not the derivative, however small $h$ is. With $h = 0.1$ , $x^{2}$ at $2$ gives a quotient of $4.1$ . That is an average rate over an interval, and the derivative is the limit, $4$ .

\frac{∣ h ∣}{h} = {\begin{cases} 1, & h > 0 \\ - 1, & h < 0 \end{cases}

Check your answer

A slope found from the limit should agree with nearby secants, with the power rule and with the picture.

Try small values of $h$ on both sides. For $x^{2}$ at $2$ , $h = 0.001$ gives $4.001$ and $h = - 0.001$ gives $3.999$ , both close to $4$ .

Once you know the power rule, use it. The derivative of $x^{2}$ is $2 x$ , which is $4$ at $x = 2$ .

Graph the tangent line. Near the point, it should hug the curve closely.

To have Chalk check the work, write the difference quotient on the first row of a Math box and each rewrite below it. Finish with the limit, as $lim (4 + h) = 4$ with $h \to 0$ . The Checker marks each rewrite ✓ or ✗ against the row above. The ✓ on $4 + h$ adds that the row above excludes $h = 0$ and this one does not. A last row $f^{'} (2) = 4$ , as courses also write it, reads ?, because the box holds no $f^{'}$ to compare it with. The limit row above it is the one Chalk checks. Finding the derivative of a square root this way, the conjugate row reads ✓ “Equivalent.”, and a wrong simplification after it reads ✗. A ✗ whose message names $h = 0$ still means the rows differ, though not only at $h = 0$ . Put $h = 1$ into both rows to see by how much.

Local linearity, and why differentiable means continuous

Write the quotient as $f^{'} (a)$ plus an error $e (h)$ that approaches $0$ . Multiplying by $h$ gives $f (a + h)$ $= f (a) + f^{'} (a) h + e (h) \cdot h$ . The last term shrinks faster than $h$ . Near $a$ , the function is its tangent line plus a much smaller correction, which is why a graph looks straight when you zoom in far enough. Linear approximation is built on this.

The same line shows why a differentiable function is continuous. As $h$ approaches $0$ , $f^{'} (a) h$ and $e (h) \cdot h$ both approach $0$ , and $f (a + h)$ approaches $f (a)$ .

An equivalent form uses a second point $x$ instead of $a + h$ . $f^{'} (a)$ is the limit of $(f (x) - f (a)) / (x - a)$ as $x$ approaches $a$ . There are even functions that are continuous everywhere and differentiable nowhere. Their graphs are corners at every scale.

f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}

Common mistakes

Setting $h = 0$ before simplifying. That gives $0 / 0$ , which is not a slope. Expand and cancel the factor $h$ while $h \neq 0$ , and only then let $h$ approach $0$ .
Expanding $f (a + h)$ as $f (a) + h$ . For $f (x) = x^{2}$ , $f (2 + h)$ is ${(2 + h)}^{2} = 4 + 4 h + h^{2}$ , not $4 + h$ . Replace every $x$ in the formula with $a + h$ .
Treating one secant slope as the derivative. With $h = 0.1$ , the quotient for $x^{2}$ at $2$ is $4.1$ , which is an estimate. The derivative is the limit, $4$ .
Assuming a continuous function has a derivative everywhere. The function $∣ x ∣$ is continuous at $0$ , but its secant slopes are $- 1$ from the left and $1$ from the right, so it has no derivative there.
Forgetting to subtract $f (a)$ . The numerator is $f (a + h) - f (a)$ . Without the subtraction, ${(2 + h)}^{2} / h$ grows without bound as $h$ approaches $0$ instead of approaching $4$ .

Key terms

Derivative: The instantaneous rate of change of a function: the limit of the average rate of change as the step shrinks to zero, when that limit exists. On a graph it is the slope of the tangent line.
Difference quotient: The average rate of change $[f (a + h) - f (a)] / h$ between $x = a$ and $x = a + h$ , with $h \neq 0$ . Its limit as $h \to 0$ is the derivative $f^{'} (a)$ , when that limit exists.
Secant line: A line through two points on a curve. Its slope is the average rate of change between them. In trigonometry, “secant” also names the function $sec θ = 1 / cos θ$ .
Tangent line: The line that matches a curve’s direction at a point, with slope equal to the derivative there. It can cross the curve elsewhere. In trigonometry, “tangent” also names $tan θ = sin θ / cos θ$ .
Limit: The value $f (x)$ approaches as $x$ gets closer and closer to a point $a$ , not counting $x = a$ itself. A limit can exist even when $f (a)$ is undefined or has a different value.
Differentiable: Having a derivative at a point. A function is continuous wherever it is differentiable, but not the other way round: $∣ x ∣$ is continuous at $x = 0$ with no derivative there.

Work through an example

Use the limit definition to find the slope of $f (x) = x^{2}$ at $x = 2$ , then write the tangent line there.

Find a derivative from the limit definition →

Find the derivative of $1 / x$ from the definition →

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