The Derivative: Slope of the Tangent Line and Rate of Change

Quick answer

The average rate of change of f from a to a + h is the slope of a secant line, (f(a + h) − f(a))/h. Letting h shrink to 0 turns secants into the tangent line and averages into an instantaneous rate: that limit is the derivative f′(a). Computed at every x it gives a new function, f′(x), also written dy/dx. A function is not differentiable at a corner, a cusp, a vertical tangent or a discontinuity; differentiability implies continuity, but not the reverse.

What you'll learn

Compute an average rate of change as the slope of a secant line
Find a derivative from the limit definition
Write the equation of a tangent line
Identify where a function is not differentiable, and why

Average rate of change

Over an interval from $a$ to $a + h$ , a function changes by $f(a + h) - f(a)$ while the input changes by $h$ . The ratio

\frac{f(a + h) - f(a)}{h}

is the average rate of change, called the difference quotient. It is the slope of the secant line through $\big(a, f(a)\big)$ and $\big(a + h, f(a + h)\big)$ .

From secant to tangent

Take $f(x) = x^2$ and $a = 1$ . As $h$ shrinks, the secant lines pivot around $(1, 1)$ and settle onto the tangent line.

y = x²
secant, h = 1
secant, h = 0.5
tangent

Secant lines settling onto the tangent

$h$	$1$	$0.5$	$0.1$	$0.01$	$0.001$
secant slope	$3$	$2.5$	$2.1$	$2.01$	$2.001$

The slopes approach $2$ , and algebra confirms it: $\tfrac{(1 + h)^2 - 1}{h} = \tfrac{2h + h^2}{h} = 2 + h \to 2$ .

The definition

The derivative of $f$ at $a$ is

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

when this limit exists. Geometrically it is the slope of the tangent line at $\big(a, f(a)\big)$ . Physically it is the instantaneous rate of change of $f$ at $a$ , measured in units of output per unit of input.

The tangent line itself is

y - f(a) = f'(a)(x - a)

Why the derivative has to be a limit

“The rate of change at an instant” sounds like a paradox. A rate compares two changes, and at a single instant nothing has changed: setting $h = 0$ in the difference quotient gives $\tfrac{0}{0}$ .

That is precisely the kind of expression limits were built for. The limit never sets $h = 0$ . It asks what the averages over ever shorter intervals are heading toward. When they close in on a single number, that number is the only sensible meaning of the instantaneous rate. The derivative is the limit of average rates, because no single average can capture an instant.

The derivative as a function

Letting the point vary gives a new function, the derivative:

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

For $f(x) = x^2$ the same algebra gives $f'(x) = 2x$ . Other common notations are

f'(x), \qquad y', \qquad \frac{dy}{dx}, \qquad \frac{d}{dx} f(x)

The $\tfrac{dy}{dx}$ notation, due to Leibniz, recalls where the derivative came from: a small change in $y$ over a small change in $x$ .

Where the derivative fails

A differentiable function is continuous: if $f'(a)$ exists, then $f(a + h) - f(a) = h \cdot \tfrac{f(a + h) - f(a)}{h} \to 0 \cdot f'(a) = 0$ . The converse is false. A function can be continuous and still have no derivative at a point:

y = |x| (corner)
y = x^(2/3) (cusp)
y = ∛x (vertical tangent)

Continuous but not differentiable at 0

Corner: the one-sided slopes differ ( $-1$ and $1$ for $|x|$ ).
Cusp: the one-sided slopes go to $+\infty$ and $-\infty$ .
Vertical tangent: the slopes go to $\infty$ from both sides.
Discontinuity: a function that is not continuous at $a$ cannot be differentiable there.

Worked examples

Common mistakes

Practice problems

Find the average rate of change of $f(x) = x^3$ from $x = 1$ to $x = 3$ .

Answer

$13$

Full solution

$\tfrac{f(3) - f(1)}{3 - 1} = \tfrac{27 - 1}{2} = 13$ .
Use the definition to find $f'(2)$ for $f(x) = 3x^2 - x$ .

Answer

$11$

Full solution

$f(2 + h) - f(2) = 3(4 + 4h + h^2) - (2 + h) - 10 = 11h + 3h^2$ . Divided by $h$ : $11 + 3h \to 11$ .
Use the definition to find $f'(x)$ for $f(x) = 5x + 2$ .

Answer

$5$

Full solution

$\tfrac{5(x + h) + 2 - (5x + 2)}{h} = \tfrac{5h}{h} = 5$ . A line has the same slope everywhere.
Find the tangent line to $y = x^2$ at $x = -1$ .

Answer

$y = -2x - 1$

Full solution

$f'(x) = 2x$ , so the slope is $-2$ and the point is $(-1, 1)$ . Then $y - 1 = -2(x + 1)$ , so $y = -2x - 1$ .
Use the definition to find $f'(x)$ for $f(x) = x^3$ .

Answer

$3x^2$

Full solution

$(x + h)^3 - x^3 = 3x^2h + 3xh^2 + h^3$ . Divided by $h$ : $3x^2 + 3xh + h^2 \to 3x^2$ .
A ball’s height is $s(t) = 20t - 5t^2$ meters after $t$ seconds. Find its velocity at $t = 1$ .

Answer

$10$ meters per second

Full solution

$s(1 + h) - s(1) = 20h - 5(2h + h^2) = 10h - 5h^2$ . Divided by $h$ : $10 - 5h \to 10$ .
Use the definition to find $f'(x)$ for $f(x) = \tfrac{1}{x + 2}$ .

Answer

$-\tfrac{1}{(x + 2)^2}$

Full solution

$\tfrac{1}{x + h + 2} - \tfrac{1}{x + 2} = \tfrac{-h}{(x + h + 2)(x + 2)}$ . Divided by $h$ : $\tfrac{-1}{(x + h + 2)(x + 2)} \to -\tfrac{1}{(x + 2)^2}$ .
Is $f(x) = |x - 2|$ differentiable at $x = 2$ ? At $x = 5$ ?

Answer

Not at $2$ ; yes at $5$ (with $f'(5) = 1$ ).

Full solution

At $2$ the graph has a corner: the one-sided slopes are $-1$ and $1$ . Near $5$ , $f(x) = x - 2$ , a line with slope $1$ .
The limit $\lim_{h \to 0} \tfrac{(2 + h)^4 - 16}{h}$ is $f'(a)$ for which function $f$ and number $a$ ?

Answer

$f(x) = x^4$ and $a = 2$

Full solution

It matches $\tfrac{f(a + h) - f(a)}{h}$ with $f(a + h) = (2 + h)^4$ and $f(a) = 16 = 2^4$ .
A student finds the tangent line to $y = x^2$ at $x = 1$ as $y - 1 = 2x(x - 1)$ . What went wrong?

Hint

Is that equation a line?

Answer

The slope must be the number $f'(1) = 2$ , not the function $2x$ . The tangent is $y = 2x - 1$ .

Full solution

$y - 1 = 2x(x - 1)$ is a parabola. A tangent line needs one fixed slope: the derivative evaluated at the point.

With $f'(1) = 2$ : $y - 1 = 2(x - 1)$ , so $y = 2x - 1$ .

Frequently asked questions

What is the derivative?

The limit of the difference quotient: f′(a) = lim as h → 0 of (f(a + h) − f(a))/h. It is the slope of the tangent line at a and the instantaneous rate of change of f there.

What is the difference between average and instantaneous rate of change?

The average rate is the change over an interval, the slope of a secant line. The instantaneous rate is the limit of those averages as the interval shrinks to a point.

How do I find the equation of a tangent line?

Find the point (a, f(a)) and the slope f′(a), then use point-slope form: y − f(a) = f′(a)(x − a).

Where is a function not differentiable?

At corners, cusps, vertical tangents and discontinuities — anywhere the difference quotient has no finite limit.

Does differentiable mean continuous?

Yes, a differentiable function is continuous. The reverse fails: |x| is continuous at 0 but has a corner there.

What to learn next

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