Chapter 6: The Meaning of the Derivative

Q: Does f'(x) = 0 always mean there is a local extremum?

No. For f(x) = x^3, we have f'(0) = 0, but x = 0 is an inflection point, not a local extremum. The condition f'(x) = 0 is necessary for a local extremum, but it is not sufficient. Check how the sign of f'(x) changes on either side of the point: positive to negative gives a local maximum, and negative to positive gives a local minimum.

Q: How can we judge the accuracy of a linear approximation?

For the linear approximation f(a+h) ≈ f(a) + f'(a)h, the error divided by h approaches 0 (the error becomes smaller faster than h). If f'' exists, the error is approximately (1/2)f''(a)h^2. The smaller h is, the more accurate the approximation. For example, when approximating √4.1 (h = 0.1), the error is very small, about 0.00015.

Meaning of the Derivative

Introduction

6.1 What the Derivative Represents: How Fast Things Change

The derivative represents the instantaneous rate at which a function changes. In this chapter, we will explore its meaning from several points of view.

The Meaning of the Derivative

$f'(x)$ represents the instantaneous rate of change of $f(x)$ at $x$.

If $f'(x) > 0$, then $f(x)$ is increasing there.
If $f'(x) < 0$, then $f(x)$ is decreasing there.
If $f'(x) = 0$, then the tangent line is horizontal there (the instantaneous rate of change is $0$).

$x$

$y$

$f'(x) \lt 0$

Decreasing

$f'(x) \gt 0$

Increasing

$f'(0) = 0$ (horizontal tangent)

$y = x^2$

Figure 1: The graph of $f(x) = x^2$ and tangent lines at three points. At $x = -1.2$, the tangent line slopes downward to the right ($f'(-1.2) = -2.4 \lt 0$), so $f$ is decreasing. At $x = 1.2$, it slopes upward to the right ($f'(1.2) = 2.4 \gt 0$), so $f$ is increasing. At $x = 0$, the tangent line is horizontal ($f'(0) = 0$), and the function changes from decreasing to increasing.

6.2 Velocity and the Derivative

One of the most basic applications of the derivative in physics is velocity.

Definition of Velocity

Suppose position $x(t)$ is given as a function of time $t$:

Average velocity: $\displaystyle \dfrac{x(t_2) - x(t_1)}{t_2 - t_1}$ (over the time interval $[t_1, t_2]$)
Instantaneous velocity: $\displaystyle v(t) = \dfrac{dx}{dt} = x'(t)$

$1$

$2$

$t$

$x$

$0.5$

$4.9$

Slope $= \dfrac{4.9}{0.5} = 9.8$

Position $x(t) = 4.9t^2$

Derivative

$1$

$2$

$t$

$v$

$9.8$

$v(1) = 9.8$

Velocity $v(t) = x'(t) = 9.8t$

Slope of the left graph's tangent line = height of the right graph

Figure 2: The distance fallen $x(t) = 4.9t^2$ [m] (left) and velocity $v(t) = 9.8t$ [m/s] (right) for free fall ($g = 9.8$), where $0 \le t \le 2$. The slope $9.8$ of the tangent line at $t = 1$ on the left equals the height $v(1) = 9.8$ at $t = 1$ on the right.

Example: Falling Motion

The distance an object falls in $t$ seconds after being released from rest (taking downward as the positive direction) is:

$$x(t) = \dfrac{1}{2}gt^2$$

Here, $g \approx 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Find the velocity

$$v(t) = \dfrac{dx}{dt} = \dfrac{d}{dt}\left(\dfrac{1}{2}gt^2\right) = \dfrac{1}{2}g \cdot 2t = gt$$

Interpretation

At $t = 0$: $v(0) = 0$ (the object starts falling from rest)
At $t = 1$ second: $v(1) = 9.8$ m/s
At $t = 2$ seconds: $v(2) = 19.6$ m/s

Velocity increases in proportion to time.

6.3 Acceleration and the Second Derivative

The rate of change of velocity is acceleration. It is the second derivative of position.

Definition of Acceleration

Given the velocity $v(t) = x'(t)$:

$$\text{acceleration} = a(t) = \dfrac{dv}{dt} = v'(t) = x''(t)$$

$t$

Position $x(t) = \tfrac{1}{2}gt^2$

$\dfrac{d}{dt}$

$t$

Velocity $v(t) = gt$

$\dfrac{d}{dt}$

$t$

Acceleration $a(t) = g$

$x(t) \;\xrightarrow{\ d/dt\ }\; x'(t) = v(t) \;\xrightarrow{\ d/dt\ }\; x''(t) = v'(t) = a(t)$

Taking the derivative of position twice gives acceleration

Figure 3: Position, velocity, and acceleration in free fall ($0 \le t \le 2$). Taking the derivative of position $x(t) = \frac{1}{2}gt^2$ once with respect to $t$ gives velocity $v(t) = gt$. Taking it once more gives acceleration $a(t) = g$ (constant). The vertical axes of all three graphs use the same scale ($0$ to $20$).

Example: Motion with Constant Acceleration

Suppose position is given by $x(t) = 3t^2 + 2t + 1$:

Velocity

$$v(t) = x'(t) = \dfrac{d}{dt}(3t^2 + 2t + 1) = 6t + 2$$

Acceleration

$$a(t) = v'(t) = x''(t) = \dfrac{d}{dt}(6t + 2) = 6$$

Interpretation

The acceleration is constant at $6$ (motion with constant acceleration).
The initial velocity is $v(0) = 2$.
The initial position is $x(0) = 1$.

6.4 Increasing, Decreasing, and Local Extrema

The derivative lets us find where a function is increasing and where it is decreasing.

How to Determine Whether a Function Is Increasing or Decreasing

If $f'(x) > 0$ on an interval, then $f(x)$ is increasing on that interval.

If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing on that interval.

Finding Local Maxima and Minima

Local maximum: If $f'(a) = 0$ and the sign of $f'(x)$ changes from positive to negative on either side of $x = a$, then $f$ has a local maximum at $x = a$.

Local minimum: If $f'(a) = 0$ and the sign of $f'(x)$ changes from negative to positive on either side of $x = a$, then $f$ has a local minimum at $x = a$.

The precise definitions of local maxima and minima, and how to handle points where $f'$ does not exist, are covered in Chapter 15.

$x$

$y$

$-1$

$1$

Local maximum $(-1,\ 2)$

$f'(-1) = 0$

$f'(1) = 0$

Local minimum $(1,\ -2)$

Increasing ↗

Decreasing ↘

Increasing ↗

$y = x^3 - 3x$

Figure 4: The graph of the example function $f(x) = x^3 - 3x$. It has a local maximum value of $2$ at $x = -1$ and a local minimum value of $-2$ at $x = 1$. The tangent line is horizontal at both points ($f'(x) = 0$). The sign of $f'(x)$ changes on either side of each point.

Example: Where $f(x) = x^3 - 3x$ Is Increasing or Decreasing

Step 1: Find the derivative.

$$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x+1)(x-1)$$

Step 2: Find the points where $f'(x) = 0$.

$$3(x+1)(x-1) = 0$$ $$x = -1 \text{ or } x = 1$$

Step 3: Check the sign on each interval.

$x$	$x < -1$	$x = -1$	$-1 < x < 1$	$x = 1$	$x > 1$
$f'(x)$	$+$	$0$	$-$	$0$	$+$
$f(x)$	↗	Local maximum	↘	Local minimum	↗

Step 4: Find the local extrema.

Local maximum value: $f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$
Local minimum value: $f(1) = (1)^3 - 3(1) = 1 - 3 = -2$

6.5 The Equation of a Tangent Line

The derivative can be used to find the equation of the tangent line at any point on a curve.

Equation of a Tangent Line

The equation of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$ is:

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

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

Derivation

The tangent line passes through $(a, f(a))$ and has slope $f'(a)$.

The equation of a line with slope $m$ through $(x_0, y_0)$ is:

$$y - y_0 = m(x - x_0)$$

Substituting $x_0 = a$, $y_0 = f(a)$, and $m = f'(a)$ gives:

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

$x$

$y$

$1$

$f'(2) = 4$

$(2,\ 4)$

$2$

$1$

$y = x^2$

Tangent line $y = 4x - 4$

Figure 5: The tangent line $y = 4x - 4$ to $y = x^2$ at $(2,\ 4)$ (from the example below). On the tangent line, when $x$ increases by $1$, $y$ increases by $f'(2) = 4$. The tangent line meets the $x$-axis at $x = 1$.

Example: The Tangent Line to $f(x) = x^2$ at $x = 2$

Step 1: Find the derivative.

$$f'(x) = 2x$$

Step 2: Find the values at $x = 2$.

$$f(2) = 2^2 = 4$$ $$f'(2) = 2 \cdot 2 = 4$$

Step 3: Write the equation of the tangent line.

$$y - 4 = 4(x - 2)$$ $$y = 4x - 8 + 4$$ $$y = 4x - 4$$

6.6 Linear Approximation

The derivative can also be used to approximate a function with a straight line.

Linear Approximation

When $x$ is close enough to $a$:

$$f(x) \approx f(a) + f'(a)(x - a)$$

This is called a linear approximation (or a first-order approximation).

$x$

$y$

$4$

$6$

$12$

Error about $0.05$

Error about $0.54$

$y = \sqrt{x}$

Tangent line $y = 2 + \tfrac{1}{4}(x - 4)$

Figure 6: $f(x) = \sqrt{x}$ and the tangent line (linear approximation) $y = 2 + \frac{1}{4}(x - 4)$ at $a = 4$. At $x = 6$, the difference between the tangent line and the curve (the error) is about $0.05$. At $x = 12$, it grows to about $0.54$. The farther $x$ is from $a$, the worse the approximation becomes.

How large is the error in this approximation? Let $h = x - a$, and define the error as

$$E(h) = f(a+h) - \bigl[f(a) + f'(a)h\bigr]$$

Dividing both sides by $h$, the definition of the derivative gives

$$\dfrac{E(h)}{h} = \dfrac{f(a+h) - f(a)}{h} - f'(a) \to 0 \quad (h \to 0)$$

This means that the error $E(h)$ approaches $0$ faster than $h$ itself (this is sometimes written as $E(h) = o(h)$). If we use a line whose slope is not $f'(a)$, then $\dfrac{E(h)}{h}$ does not approach $0$. This is why the tangent line is called the "best linear approximation."

$x$

$y$

$a = 4$

$a + h = 9$

$h = 5$

$(a,\ f(a)) = (4,\ 2)$

$f'(a)\,h = 1.25$

Error $0.25$

$f(a) + f'(a)h = 3.25$

$f(a+h) = 3$

$y = \sqrt{x}$

Figure 7: The parts of a linear approximation ($f(x) = \sqrt{x}$, $a = 4$, $h = 5$). The value on the tangent line is $f(a) + f'(a)h = 2 + \frac{1}{4} \cdot 5 = 3.25$. The difference $0.25$ between this and the actual value $f(9) = 3$ is the error. As $h$ gets smaller, the error shrinks much faster than $h$ does (when $h = 1$, the error is about $0.014$).

Example: Approximating $\sqrt{4.1}$

Use $f(x) = \sqrt{x}$, $a = 4$, and $x = 4.1$.

Step 1: Find the values we need.

$$f(4) = \sqrt{4} = 2$$ $$f'(x) = \dfrac{1}{2\sqrt{x}} \Rightarrow f'(4) = \dfrac{1}{2\sqrt{4}} = \dfrac{1}{4}$$

Step 2: Apply the linear approximation formula.

$$\sqrt{4.1} \approx f(4) + f'(4)(4.1 - 4)$$ $$= 2 + \dfrac{1}{4} \times 0.1$$ $$= 2 + 0.025 = 2.025$$

Check: A calculator gives $\sqrt{4.1} \approx 2.0248...$

The linear approximation $2.025$ is very accurate.

6.7 Chapter Summary

Summary: The Meaning of the Derivative

Viewpoint	Meaning of $f'(x)$
Geometry	Slope of the tangent line
Physics	Instantaneous rate of change (such as velocity)
Calculus	A way to tell where a function is increasing or decreasing
Approximation	Coefficient in a linear approximation

Frequently Asked Questions

What is the difference between a local extremum and an absolute maximum or minimum?

A local extremum is a maximum or minimum when values are compared only near that point. An absolute maximum or minimum is the greatest or least value over the entire domain, so a local extremum may not be an absolute maximum or minimum. For example, an absolute maximum can occur at an endpoint of an interval.

Does $f'(x) = 0$ always mean there is a local extremum?

No. For $f(x) = x^3$, we have $f'(0) = 0$, but $x = 0$ is an inflection point, not a local extremum. The condition $f'(x) = 0$ is necessary for a local extremum, but it is not sufficient. Check how the sign of $f'(x)$ changes on either side of the point: positive to negative gives a local maximum, and negative to positive gives a local minimum.

How can we judge the accuracy of a linear approximation?

For the linear approximation $f(a+h) \approx f(a) + f'(a)h$, the error divided by $h$ approaches $0$ (the error becomes smaller faster than $h$). If $f''$ exists, the error is approximately $\dfrac{1}{2}f''(a)h^2$. The smaller $h$ is, the more accurate the approximation. For example, when approximating $\sqrt{4.1}$ ($h = 0.1$), the error is very small, about $0.00015$.

Exercises

Exercise 1

An object's position is given by $x(t) = t^3 - 6t^2 + 9t + 2$:

Find the velocity $v(t)$.
Find the acceleration $a(t)$.
Find the times when the velocity is $0$.

Exercise 2

For $f(x) = x^3 - 12x + 5$:

Find $f'(x)$.
Find the local maximum and local minimum values of $f(x)$.

Exercise 3

Find the equation of the tangent line to $f(x) = x^3$ at $x = 1$.

Exercise 4

Use a linear approximation to find an approximate value of $\sqrt{9.2}$.

Show the answers

Answer to Exercise 1

$v(t) = x'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$
$a(t) = v'(t) = 6t - 12$
When $v(t) = 0$, we have $(t-1)(t-3) = 0$, so $t = 1$ or $t = 3$.

Answer to Exercise 2

$f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x+2)(x-2)$
$f'(x) = 0$ at $x = -2, 2$.
- At $x = -2$: $f(-2) = (-2)^3 - 12(-2) + 5 = -8 + 24 + 5 = 21$ (local maximum value)
- At $x = 2$: $f(2) = 2^3 - 12(2) + 5 = 8 - 24 + 5 = -11$ (local minimum value)

Answer to Exercise 3

Since $f'(x) = 3x^2$, we have $f'(1) = 3$.

$f(1) = 1$

Tangent line: $y - 1 = 3(x - 1)$, so $y = 3x - 2$.

Answer to Exercise 4

Let $f(x) = \sqrt{x}$, $a = 9$, and $x = 9.2$.

$f(9) = 3$, and since $f'(x) = \dfrac{1}{2\sqrt{x}}$, we have $f'(9) = \dfrac{1}{6}$.

$\sqrt{9.2} \approx 3 + \dfrac{1}{6}(0.2) = 3 + \dfrac{0.2}{6} = 3 + \dfrac{1}{30} \approx 3.0333$

(The actual value is about $3.0332$.)