AP®MathsCollege BoardCalculus BCStudy GuidesUnit 2: Differentiation Definition & Fundamental PropertiesDefinition of DifferentiationEstimating the Derivative at a Point

Estimating the Derivative at a Point (College Board AP® Calculus BC): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 4 May 2026

Estimating derivatives at a point using a graph

How can I estimate a derivative at a point using points on the graph?

The coordinates of the points that lie on the graph of a function can be used to estimate the derivative at a point
Recall that the derivative of $f (x)$ at the point where $x = a$ , denoted as $f^{'} (a)$ ,
- is equal to the slope of the tangent to the graph of $f (x)$ at $x = a$
To approximate the slope of the tangent to the graph of $f (x)$ at $x = a$ :
- Find the slope of line segments joining nearby points that lie on the graph
The function must be continuous and differentiable within the relevant interval for this method to be valid
Consider the graph of $f (x)$ below, where points $A, B, C, D, E$ are labeled with their coordinates

A graph of the function f(x) with labeled points: A (1, -6), B (2, -6), C (3, -4), D (4, 0), and E (5, 6). The graph follows a curved path.

To estimate the derivative at point $C$ you can find the slope of the nearest secant line
- Finding the slope between $B$ and $D$
  - $\frac{0 - - 6}{4 - 2} = \frac{6}{2} = 3$

How can I estimate a derivative at a point using a tangent line?

To estimate the derivative of a function at a point
- Draw a tangent to the curve at the point
- Pick two points on the line
- Calculate the slope

Finding the slope of tangent — **Example of estimating the derivative at a point using a tangent line**

Estimating derivatives at a point using a table

How can I estimate a derivative at a point using a table?

The same method can be used as for a graph, but with a table of values instead
Recall that the derivative of $f (x)$ at the point where $x = a$ , denoted as $f^{'} (a)$ ,
- is equal to the slope of the tangent to the graph of $f (x)$ at $x = a$
To approximate the slope of the tangent to the graph of $f (x)$ at $x = a$ :
- Find the average rate of change using points on either side of $x = a$
Consider the table of values below for the function $g$
- $g$ is a continuous and differentiable function within this interval
- This must be true to use this method

$x$	$g (x)$
1	-16
3	-24
5	-24
7	-16
9	0
11	24

To find an estimate for $g^{'} (7)$ :
- Find the average rate of change between $x = 5$ and $x = 9$
- $\frac{0 - (- 24)}{9 - 5} = 6$
To find an estimate for $g^{'} (6)$ :
- Find the average rate of change between $x = 5$ and $x = 7$
- $\frac{- 16 - (- 24)}{7 - 5} = 4$

Examiner Tips and Tricks

In an exam, the question might ask you to estimate an instantaneous rate of change and give you a table with lots of values. According to the AP Calculus Chief Reader Reports, you are expected to estimate the derivative value using a difference quotient by drawing from the data in the table that most tightly bounds the specified point. This means you need to choose the two closest values. In every past FRQ, the test designers deliberately chose a target value that falls exactly in the middle of those two closest bounding points.

Worked Example

$x$	1	2	3	4	5
$f (x)$	0	4.5	7.2	9.1	12.4

Selected values of the differentiable function $f (x)$ are given in the table above.

Use the data in the table to approximate $f' (2.5)$ . Show the computations that lead to your answer.

Answer:

Identify the value of the function at the values closest to $x = 2.5$

$f (2) = 4.5$ and $f (3) = 7.2$

Calculate the average rate of change between these two points

$\begin{array}{rcl} f' (2.5) & \approx & \frac{f (3) - f (2)}{3 - 2} \\ = & \frac{7.2 - 4.5}{1} \\ = & 2.7 \end{array}$

$\begin{array}{rcl} f' (2.5) & \approx & 2.7 \end{array}$

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