2.3 - Estimating Derivatives of a Function at a Point

Introduction

Graphing calculators can compute derivatives at specific points, even for functions you don’t yet know how to differentiate by hand. That’s incredibly useful for calculator-allowed portions of exams and homework.

Important: You need a graphing calculator for this lesson. If you don’t have a TI-84 or TI-83, look up instructions specific to your calculator model.

Calculator: Finding Derivatives at a Point

The nDeriv Function (TI-84)

Your calculator can estimate the derivative at any specific

x

-value.

Steps:

Press MATH

Scroll down to option 8: nDeriv, or just press 8

The format appears: nDeriv(function, variable, value)

Syntax:


nDeriv(f(x), X, a)

where f(x) is your function, X is the variable, and a is the

x

-value where you want the derivative.

Result: this gives you

f'(a)

, the slope of the tangent line at

x = a

Example 1: Basic Derivative Calculation

Find

f'(3)

f(x) = e^x \cos(x)

Calculator steps:

Press MATH → 8

Enter nDeriv(e^(X)cos(X), X, 3)

Press ENTER

Result:

-22.719

Interpretation: The slope of the tangent line to

f(x)

x = 3

is about

-22.719

Answer:

f'(3) \approx -22.719

Example 2: Function with a Square Root

Find the derivative of

g(x) = \sqrt{x^2 + 4}

x = 5

Calculator steps:

MATH → 8

Enter nDeriv(√(X² + 4), X, 5)

Press ENTER

Result:

0.928

Answer:

g'(5) \approx 0.928

, rounded to three decimal places

Example 3: Logarithmic Function

Find

h'(2)

h(x) = \ln\left(\frac{3}{x + 1}\right)

Calculator steps:

MATH → 8

Enter nDeriv(ln(3/(X + 1)), X, 2)

Press ENTER

Result:

-0.333

Answer:

h'(2) \approx -0.333

Using the Calculator for Tangent Line Equations

Recall: Point-Slope Form

y - y_1 = m(x - x_1)

For tangent lines:

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

What you need: 1. The

x

-value:

a

2. The

y

-value:

f(a)

, from evaluating the function 3. The slope:

f'(a)

, from the calculator derivative

Example 4: Complete Tangent Line Problem

Find the equation of the tangent line to

f(x) = x \ln(x)

x = 2

Step 1: Find the $y$ -value

f(2) = 2 \ln(2) \approx 1.386

Step 2: Find the slope (the derivative at $x = 2$ )

Calculator: nDeriv(X*ln(X), X, 2) gives

f'(2) \approx 1.693

Step 3: Write the equation

y - 1.386 = 1.693(x - 2)

Answer:

y - 1.386 = 1.693(x - 2)

Note: Unless you’re asked to solve for

y

, leave it in point-slope form.

Example 5: Another Tangent Line

Write the equation of the tangent line to

g(x) = \frac{e^x}{x}

x = 1

Find $g(1)$ :

g(1) = \frac{e^1}{1} = e \approx 2.718

Find $g'(1)$ :

Calculator: nDeriv(e^(X)/X, X, 1) gives

g'(1) \approx 0

Write the equation:

y - 2.718 = 0(x - 1) \quad \Rightarrow \quad y - 2.718 = 0

Answer:

y = 2.718

, a horizontal tangent line.

Estimating Derivatives from Tables

When you have a table of function values but no formula, you can estimate the derivative using nearby points.

Method: Average Rate of Change

To estimate

f'(c)

where

c

sits between

a

and

b

f'(c) \approx \frac{f(b) - f(a)}{b - a}

Key principle: use the two

x

-values closest to and surrounding your target point.

Example 6: Basic Table Estimation

The function

h

is differentiable. Use the table to estimate

h'(5)

$x$	2	5	8	11
$h(x)$	10	18	30	46

We want

h'(5)

, and the closest surrounding values are at

x = 2

and

x = 8

h'(5) \approx \frac{h(8) - h(2)}{8 - 2} = \frac{30 - 10}{6} = \frac{20}{6} = \frac{10}{3}

Answer:

h'(5) \approx 3.333

Example 7: Table with Units

The table shows distance

d(t)

in kilometers after

t

hours. Estimate

d'(3)

$t$ (hours)	1	3	5	7
$d(t)$ (km)	50	140	250	380

Use the surrounding values at

t = 1

and

t = 5

d'(3) \approx \frac{d(5) - d(1)}{5 - 1} = \frac{250 - 50}{4} = \frac{200}{4} = 50

Units: since

d

is in kilometers and

t

is in hours,

d'

is in kilometers per hour.

Answer:

d'(3) \approx 50

km/h

Interpretation: at

t = 3

hours, the speed is about 50 km/h.

Example 8: Rate of a Rate (Second Derivative Concept)

The function

v(t)

gives velocity in m/s at time

t

in seconds. Estimate

v'(4)

$t$ (seconds)	0	4	8
$v(t)$ (m/s)	20	35	44

v'(4) \approx \frac{v(8) - v(0)}{8 - 0} = \frac{44 - 20}{8} = \frac{24}{8} = 3

Units: -

v

is in meters per second (m/s) -

t

is in seconds (s) - so

v'

is in (m/s) per second, which is m/s²

Answer:

v'(4) \approx 3

m/s²

Interpretation: at

t = 4

seconds, the acceleration is about 3 m/s².

Understanding Units with Derivatives

When the Function is Already a Rate

f(x)

represents a rate, like gallons per minute, then

f'(x)

represents a rate of change of a rate.

Original function	Derivative
meters	meters/second
miles/hour	miles/hour² or (miles/hour)/hour
gallons/second	gallons/second² or (gallons/second)/second

Shorthand: “per second per second” becomes “per second squared.”

Example 9: Flow Rate

Water flows from a tank at a rate

R(t)

gallons per minute, where

t

is time in minutes.

$t$ (min)	0	5	10
$R(t)$ (gal/min)	40	32	20

Estimate

R'(5)

and give the units.

R'(5) \approx \frac{R(10) - R(0)}{10 - 0} = \frac{20 - 40}{10} = \frac{-20}{10} = -2

Units: -

R

is in gallons per minute -

t

is in minutes - so

R'

is in (gallons/minute) per minute, which is gallons per minute squared

Answer:

R'(5) \approx -2

gal/min²

Interpretation: at

t = 5

minutes, the flow rate is decreasing at 2 gallons per minute squared.

Estimating Derivatives from Graphs

If all you have is a graph of

f

, then

f'(a)

is the slope of the tangent line at

x = a

. You can estimate it by reading two points off that tangent line.

Draw the tangent line at $x = a$ (or use the one given). It should touch the curve at that point and match its direction there.

Pick two points on the tangent line that are easy to read, like spots where it crosses grid lines.

Find rise over run between those two points.

Quick sign check: if the graph is going up at

x = a

, then

f'(a)

is positive. If it’s going down,

f'(a)

is negative. At a smooth peak or valley, where the tangent line is horizontal,

f'(a) = 0

Example 10: Reading a Slope from a Graph

The graph shows

y = f(x)

and the line tangent to it at

x = 2

. Estimate

f'(2)

, and decide whether

f'(5)

is positive or negative.

2.3 - Estimating Derivatives of a Function at a Point

Step 1: Find two easy points on the tangent line. It passes through

(2, 3)

, the point of tangency, and

(5, 5)

Step 2: Find rise over run.

f'(2) \approx \frac{5 - 3}{5 - 2} = \frac{2}{3} \approx 0.667

Step 3: Check the sign. The graph is rising at

x = 2

, so a positive slope makes sense.

For

f'(5)

: the graph is falling at

x = 5

(it’s past the peak), so

f'(5)

is negative.

Answer:

f'(2) \approx \frac{2}{3}

, or about

0.667

, and

f'(5) < 0

Important Notes About Calculator Derivatives

Accuracy

The calculator’s nDeriv function gives you an estimate, not an exact answer. That said, it’s usually accurate to 4 or 5 decimal places, which is way more than enough for our purposes.

Watch out: nDeriv estimates the slope using points on both sides of

x = a

, so it can give you a number even where the derivative doesn’t exist. For example, nDeriv(abs(X), X, 0) gives

0

, but

f(x) = |x|

has a sharp corner at

x = 0

, so

f'(0)

doesn’t exist (Topic 2.4). If the graph has a corner, cusp, jump, or vertical tangent at that point, don’t trust the calculator’s answer.

When to Use It

On calculator-allowed portions of exams

For functions you don’t yet know how to differentiate

To check your work while you’re learning new derivative rules

When NOT to Use It

✗ On no-calculator portions

✗ Later in the year, when you’ll be able to do these by hand faster

✗ On exam questions that require you to show the process, not just the answer

Practice Problems

Problem 1

Use your calculator to find

f'(1)

f(x) = x^2 \sin(x)

Answer: nDeriv(X²sin(X), X, 1) gives about

2.223

Problem 2

Write the equation of the tangent line to

g(x) = \sqrt{x + 3}

x = 6

Answer:

g(6) = 3

and

g'(6) \approx 0.167

, so the equation is

y - 3 = 0.167(x - 6)

Problem 3

Estimate

h'(4)

from the table:

$x$	2	4	6
$h(x)$	15	23	28

Answer:

h'(4) \approx \frac{28 - 15}{6 - 2} = \frac{13}{4} = 3.25

Problem 4

Temperature

T(t)

in degrees Celsius after

t

minutes:

$t$	0	3	6
$T(t)$	100	85	73

Estimate

T'(3)

with units.

Answer:

T'(3) \approx \frac{73 - 100}{6 - 0} = \frac{-27}{6} = -4.5 \text{ °C/min}

Key Takeaways

Calculator command: MATH → 8 for nDeriv on the TI-84

Syntax: nDeriv(function, X, value) gives the derivative at that point

The calculator gives approximate values, accurate to several decimal places

For tangent lines, you need the point $(a, f(a))$ and the slope $f'(a)$

From tables, estimate $f'(c)$ using $\frac{f(b) - f(a)}{b - a}$ with surrounding points

Use the closest surrounding points for the best estimate

From a graph, estimate $f'(a)$ as the slope of the tangent line, using two easy-to-read points on it

Pay attention to units. Derivative units include “per”

Rate of a rate: “per second per second” becomes “per second squared”

This method only works if the function is differentiable, meaning smooth and continuous

Round to three decimal places for final answers

The calculator is a powerful tool, but you’ll learn to do these by hand soon

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