Interpreting the Derivative

Main idea

One number, three readings: $f'(a)$ is the slope of the tangent line, how fast $f$ is changing right then, and, if $f$ is position, the velocity. Its sign tells you whether $f$ is going up or down, and its units tell you what it means.

Tangent line at x = a

y = f(a) + {\htmlData{sym=0}{f'(a)}}(x - a)

What the symbols mean

$f'(a)$
Slope at a
Positive: the function is increasing there. Negative: decreasing. Zero: momentarily flat.

Explanation

Interpreting a derivative simply means figuring out what the number $f'(a)$ is telling you about the real situation. All we're really saying is that the slope at a point, the rate of change at an instant and the velocity of a moving object are the same number seen three ways. The tangent line formula $y = f(a) + f'(a)(x - a)$ looks like a mouthful, but it's just point-slope form: $f(a)$ (the height where you start), $f'(a)$ (the slope there) and $x - a$ (how far you've moved from $a$ ). Nothing new is being computed here; it's about reading what you already know how to find.

Units. The derivative is rise over run, so its units are output units per input unit. If $V(t)$ is liters and $t$ is minutes, $V'(t)$ is liters per minute. When you're not sure what a derivative means in a word problem, reading the units off usually tells you.

Sign. A positive slope goes uphill, a negative one goes downhill, and that's all the sign is saying:

$f'(a) > 0$ : $f$ is increasing near $a$ .
$f'(a) < 0$ : $f$ is decreasing near $a$ .
$f'(a) = 0$ : horizontal tangent. $f$ may be at a peak, a valley, or just pausing for a moment. We'll use this to find maximums and minimums in the next unit.

Motion. If $s(t)$ is position, $s'(t)$ is velocity, which is just the instantaneous velocity from the very first lesson. Its sign is the direction of motion, and the object is at rest when $s'(t) = 0$ . The size $|s'(t)|$ , without the direction, is speed.

Marginal quantities. In economics, $C'(x)$ is called marginal cost. Since $C'(x)$ is extra cost per extra unit, it's roughly the cost of producing one more unit when you're already making $x$ .

See it

This is a cart's position, s(t) = t³ − 6t² + 9t, and the slope of the tangent at each moment is its velocity. As you can see, it starts out rolling forward at 9 m/s, stops at t = 1 to turn around, backs up until t = 3 (fastest at t = 2, going 3 m/s backward), then stops again and heads forward for good.

Worked examples

Example 1

An object moves with $s(t) = t^3 - 6t^2 + 9t$ meters. When is it moving left, and when is it at rest?

1
The velocity is $s'(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

$P(t)$ is a city's population in thousands, $t$ years after 2020, and $P'(5) = 2.3$ . What does that mean?

Problem 2+5 XP

Find the tangent line to $f(x) = x^3$ at $x = 2$ .

Problem 3+5 XP

Is $f(x) = x^3 - 6x^2$ increasing or decreasing at $x = 3$ ?

Problem 4+5 XP

For $s(t) = t^3 - 6t^2 + 9t$ , find the velocity at $t = 2$ .

Problem 5+5 XP

When is an object with $s(t) = t^2 - 8t + 3$ at rest?

Problem 6+5 XP

A company's profit from selling $x$ units is $P(x)$ dollars, with $P(500) = 3000$ and $P'(500) = 10$ . Estimate the profit from selling $503$ units.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Two tangent lines to $y = x^2$ pass through the point $(2, -5)$ , which lies below the parabola. Find the $x$ -coordinates of their points of tangency.

Where it's used

A speedometer shows $s'(t)$ ; a pharmacist's dose-response curve uses $C'(t)$ to see how fast a drug level is rising; an economist uses $C'(x)$ to decide whether making one more unit is worth it.

Position, Velocity, AccelerationCalc I · Physics Maximizing ProfitCalc I · Economics & Finance

The Definition of the Derivative Differentiation Formulas