Estimating Derivatives

When

f

can be evaluated anywhere, two quotients are standard. The one-sided quotient

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

, with

h > 0

, uses

a

and a point to its right. The centered quotient

\frac{f(a + h) - f(a - h)}{2h}

uses points on both sides. For the same

h

, the centered quotient is usually the closer estimate. On a curving graph the one-sided secant tilts toward the side it looks at, and the centered secant balances the two sides.

One-sided: the secant from a to a + h (blue) is steeper than the tangent at a (dashed orange).

Centered: the secant from a − h to a + h is nearly parallel to the tangent at a.

From a table

Example 1

A balloon is being inflated. Its radius

r(t)

, in centimeters, is measured at the times

t

, in seconds, given in the table, and

r

is differentiable.

$t$ (seconds)	$r(t)$ (cm)
$0$	$6.0$
$3$	$7.8$
$5$	$8.6$
$9$	$9.8$
$14$	$11.0$

(a) Estimate

r'(4)

(b) Estimate

r'(5)

r'(14)

(a) The nearest data on either side of

4

are at

3

and

5

r'(4) \approx \frac{8.6 - 7.8}{5 - 3} = 0.4

cm per second.

(b) Here

5

is itself a data point, so it can sit at the end of the interval. The shortest choice is

[3, 5]

, which gives

r'(5) \approx \frac{8.6 - 7.8}{5 - 3} = 0.4

cm per second. The intervals

[5, 9]

and

[3, 9]

give

0.3

and

\frac{1}{3}

, which are also reasonable estimates.

(c) There are no data after

14

, so use the last interval:

r'(14) \approx \frac{11.0 - 9.8}{14 - 9} = 0.24

cm per second.

From a formula, with a calculator

Example 2

Let

f(x) = \sqrt{1 + x^3}

. Estimate

f'(2)

with the one-sided and centered difference quotients for

h = 0.1

and

h = 0.01

, and compare them with the calculator's numerical derivative.

Here

f(2) = 3

. To four decimal places,

$h$	one-sided	centered
$0.1$	$2.0328$	$1.9994$
$0.01$	$2.0033$	$2.0000$

For example, with

h = 0.1

the centered quotient is

\begin{aligned} &\frac{f(2.1) - f(1.9)}{0.2} \\ &\qquad \approx \frac{3.203280 - 2.803391}{0.2} \\ &\qquad \approx 1.999445. \end{aligned}

A TI-84 gives

1.999999944

. The exact value is

2

, which the conjugate shows:

\begin{aligned} &\frac{f(2 + h) - 3}{h} \\ &\qquad = \frac{(2 + h)^3 - 8}{h\left(\sqrt{1 + (2 + h)^3} + 3\right)} \\ &\qquad = \frac{12 + 6h + h^2}{\sqrt{1 + (2 + h)^3} + 3} \\ &\qquad \to \frac{12}{6} = 2. \end{aligned}

The centered quotient with

h = 0.1

is already closer than the one-sided quotient with

h = 0.01

From a graph

What a numerical derivative misses

Example 4

A calculator gives

0

as the numerical derivative of

f(x) = |x|

x = 0

. Show that

f'(0)

doesn't exist.

The centered quotient is

\frac{|h| - |-h|}{2h} = 0

for every

h \ne 0

, which is the calculator's answer. The definition uses

\frac{f(0 + h) - f(0)}{h}

with

h

of either sign, and its value depends on that sign:

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

The one-sided limits are

1

and

-1

, so

f'(0)

doesn't exist. The graph has a corner at the origin, and the centered secant averages the two slopes into a horizontal line.

y = ∣x∣: the centered secant through (−h, h) and (h, h) is horizontal for every h, but no tangent line exists at the corner.