Lesson 3 of 4Rates of Change & Limits

Unit 1 · Lesson 3 of 4

The Derivative from First Principles

The difference quotient (f(a + h) − f(a))/h is the slope of a secant, the average rate of change from a to a + h. Its limit as h → 0 is the slope of the tangent, the instantaneous rate of change f′(a). Simplify the quotient algebraically first, and the limit becomes exact.

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Key things to remember

1Difference quotient: $\frac{f ( a + h ) - f ( a )}{h}$ , the slope of the secant from $x = a$ to $x = a + h$ (an average rate of change).
2Derivative at a point: $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h},$ the slope of the tangent (an instantaneous rate of change).
3You cannot substitute $h = 0$ directly ( $\frac{0}{0}$ ); expand, simplify and cancel the factor of $h$ first.
4For $f (x) = x^{2}$ at $x = 3$ : the quotient simplifies to $6 + h$ , so $f^{'} (3) = 6$ .
5 $f^{'} (a)$ has units of (output units) per (input unit), such as m/s.

The first two lessons gave you two tools. Secants through a point estimate the slope of the tangent, and limits describe the value those estimates approach. Put them together and you get the most important definition in the course. It turns the guesswork of a table of shrinking intervals into an exact answer, and it is called finding the derivative from first principles, because it uses nothing but the definition of slope and the idea of a limit.

Fig 1The curve f(x) = 0.2x² + 0.5 with the points (a, f(a)) and (a + h, f(a + h)) at a = 1.5, the run h and the rise f(a + h) − f(a) marked. Secants for h = 3, 2 and 1 have slopes 1.2, 1.0 and 0.8 and pivot toward the dashed tangent, whose slope is 0.6.

\frac{f ( a + h ) - f ( a )}{h} = \frac{rise}{run} = average rate of change from x = a to x = a + h

The difference quotient: the slope of the secant through (a, f(a)) and (a + h, f(a + h)). The run is (a + h) − a = h.

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

The derivative of f at x = a: the slope of the tangent, and the instantaneous rate of change at x = a, provided this limit exists.

The symbol $f^{'} (a)$ is read " $f$ prime of $a$ ". It carries the units of a rate: if $s (t)$ is a position in metres and $t$ is in seconds, then $s^{'} (3)$ is a velocity in m/s at the instant $t = 3$ . If $V (t)$ is a volume in litres and $V^{'} (5) = - 3$ , then 5 minutes in, the volume is falling at 3 L/min.

Reading a limit statement

What does $lim_{h \to 0} \frac{f ( 4 + h ) - f ( 4 )}{h} = 8$ say about the graph of $f (x) = x^{2}$ ? The quotient is the slope of a secant from the point $(4, 16)$ to a nearby point, and the limit is where those slopes end up. So the tangent to $y = x^{2}$ at $(4, 16)$ has slope $8$ : at that instant, $y$ is increasing $8$ times as fast as $x$ . (Check: the secant slopes are $8.1$ , $8.01$ and $8.001$ for $h = 0.1$ , $0.01$ and $0.001$ .) For any function $y = f (x)$ , the same statement says that the instantaneous rate of change at $x = 4$ is $8$ , which is $f^{'} (4) = 8$ .

Substitute first, or simplify first?

h	f(3 + h)	f(3 + h) − f(3)	Difference quotient
$1$	$16$	$7$	$7$
$0.1$	$9.61$	$0.61$	$6.1$
$0.01$	$9.0601$	$0.0601$	$6.01$
$0.001$	$9.006001$	$0.006001$	$6.001$

For f(x) = x² at x = 3, substituting each h into the difference quotient directly. Here f(3) = 9.

\frac{( 3 + h ) ^{2} - 9}{h} = \frac{9 + 6 h + h ^{2} - 9}{h} = \frac{6 h + h ^{2}}{h} = 6 + h, h \neq = 0

Simplifying first: expand, cancel the 9s, then divide every term by h.

The simplified form $6 + h$ reproduces every row of the table instantly: $h = 1$ gives $7$ , $h = 0.1$ gives $6.1$ , and so on. But it does far more. In the original quotient you can never put $h = 0$ , because the result is $\frac{0}{0}$ . In $6 + h$ there is no division by $h$ left, so the limit is plain to see: $lim_{h \to 0} (6 + h) = 6$ . The table only suggests $6$ ; the algebra proves it.

First principles in four steps

(1) Find

f (a)

. (2) Expand

f (a + h)

and collect like terms. (3) Subtract

f (a)

: the terms without

h

cancel, so every remaining term has a factor of

h

; divide each term by

h

. (4) Let

h \to 0

in the simplified expression.

Do not cancel h too early

You can only divide out

h

after you expand and subtract. In

\frac{( 3 + h ) ^{2} - 9}{h}

, the

h

in the denominator is not a factor of the whole numerator until you expand and subtract, so cancelling it against the

h

inside

(3 + h)^{2}

is wrong. If, after subtracting, some term has no

h

in it, look for an arithmetic error: the constant terms must cancel.

VideoMario uses the limit of the difference quotient to find the slope of the tangent at one point, simplifying algebraically before letting h go to 0. His example is a square root function, so the simplifying step is rationalizing the numerator.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

f (x) = x^{2}

x = 3

, compute the difference quotient for

h = 1, 0.1, 0.01, 0.001

, then simplify it and find

f^{'} (3)

Example 2

Use first principles to find the slope of the tangent to

f (x) = 2 x^{2} - 5 x + 1

x = 3

Example 3

Use first principles to find

f^{'} (2)

for

f (x) = x^{3}

Example 4

Use first principles to find

f^{'} (2)

for

f (x) = \frac{1}{x}

Where marks get lost

Writing $f (a + h)$ as $f (a) + h$ . For $f (x) = x^{2}$ , $f (3 + h) = (3 + h)^{2} = 9 + 6 h + h^{2}$ , not $9 + h$ .
Expanding $(3 + h)^{2}$ as $9 + h^{2}$ . The middle term $6 h$ is the one that survives the limit, so dropping it loses the answer.
Substituting $h = 0$ before simplifying, getting $\frac{0}{0}$ , and concluding the derivative is $0$ or undefined.
Forgetting to distribute a negative or a coefficient: $- 5 (3 + h) = - 15 - 5 h$ , not $- 15 + h$ or $- 15 - h$ .
Stopping at the simplified quotient and giving $7 + 2 h$ as the answer. The derivative at $x = a$ is a number: the final step must actually let $h \to 0$ .

Flashcards 7

Difference quotient

Definition of

f^{'} (a)

What

f^{'} (a)

means on the graph

What

f^{'} (a)

means as a rate

+ 3 more

Practice

Use first principles to find the slope of the tangent to

f (x) = x^{2} - 3 x

x = 5

For a function

f

lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = - 3

. What does this tell you?

For

f (x) = x^{2}

x = 3

, the difference quotient simplifies to

6 + h

. What is the slope of the secant from

x = 3

x = 3.05

Why can you not find

lim_{h \to 0} \frac{( 3 + h ) ^{2} - 9}{h}

by substituting

h = 0

right away?

Put the steps for finding

f^{'} (a)

from first principles in order.

1Subtract $f (a)$ , so the terms without $h$ cancel
2Let $h \to 0$ in the simplified expression
3Divide each remaining term by $h$
4Expand $f (a + h)$ and collect like terms

Use first principles to find

f^{'} (- 1)

for

f (x) = 4 - x^{2}

Match each expression to what it represents for the graph of

y = f (x)

h

f (a + h) - f (a)

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

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

What is the name for

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

, written

f^{'} (a)

A ball's height is

s (t) = 30 t - 5 t^{2}

metres after

t

seconds. Use first principles to find its velocity at

t = 2

, in m/s.

m/s

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