AP Calculus BC 2.3 Estimating Derivatives of a Function at a Point Study Notes - New Syllabus
AP Calculus BC 2.3 Estimating Derivatives of a Function at a Point Study Notes- New syllabus
AP Calculus BC 2.3 Estimating Derivatives of a Function at a Point Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Derivatives allow us to determine rates of change at an instant by applying limits to knowledge about rates of change over intervals.
Key Concepts:
- Estimating Derivatives of a Function at a Point
- Using Technology to Estimate or Calculate Derivatives
Estimating Derivatives of a Function at a Point
Estimating Derivatives of a Function at a Point
The derivative of a function at a specific point \( x = a \) can be estimated using nearby function values from a table or a graph. The goal is to approximate the slope of the tangent line at that point using the slope of a nearby secant line.
Numerical Estimation from a Table:
The derivative can be estimated using the symmetric difference quotient if values on both sides of the point are available:
\( f'(a) \approx \dfrac{f(a + h) – f(a – h)}{2h} \)
If values are only available on one side, use a one-sided difference:
- Forward difference: \( f'(a) \approx \dfrac{f(a + h) – f(a)}{h} \)
- Backward difference: \( f'(a) \approx \dfrac{f(a) – f(a – h)}{h} \)
Graphical Estimation:
From a graph, the derivative at a point is approximated by the slope of the tangent line. This can be done by drawing a tangent line at the point and estimating its rise over run:
\( f'(a) \approx \dfrac{\text{Change in } y}{\text{Change in } x} = \dfrac{\Delta y}{\Delta x} \)
This method is particularly useful when you don’t have a function rule but only data or a visual graph.
Example:
Use the table below to estimate \( f'(2) \). Assume the values are from a smooth function.
| x | f(x) |
|---|---|
| 1.8 | 3.45 |
| 2.0 | 3.80 |
| 2.2 | 4.12 |
▶️Answer/Explanation
To estimate \( f'(2) \), we use symmetric difference quotient:
\( f'(2) \approx \frac{f(2.2) – f(1.8)}{2.2 – 1.8} = \frac{4.12 – 3.45}{0.4} = \frac{0.67}{0.4} = 1.675 \)
So, the estimated derivative \( f'(2) \approx 1.675 \)
Example:
The graph below shows a smooth function \( f(x) \).
Estimate the derivative \( f'(1) \) by determining the slope of the tangent line to the curve at \( x = 1 \).
▶️Answer/Explanation
To estimate \( f'(1) \), draw a tangent line to the curve at \( x = 1 \) and choose two points on the tangent line to find the slope.
Suppose from the graph the tangent line passes through the points \( (0, 1.5) \) and \( (2, 4.5) \), then:
\( f'(1) \approx \frac{4.5 – 1.5}{2 – 0} = \frac{3}{2} = 1.5 \)
Estimated Derivative: \( f'(1) \approx 1.5 \)
Using Technology to Estimate or Calculate Derivatives
Using Technology to Estimate or Calculate Derivatives
Technology (like graphing calculators, graphing software, or computer algebra systems) can be used to calculate or estimate the value of a derivative of a function at a specific point.
There are two main approaches:
- Numerical Derivative: Many calculators and software systems (e.g., Desmos, GeoGebra, TI calculators, WolframAlpha) provide a function like
nDerivorderivative(f, a)to compute \( f'(a) \) numerically. - Graphical Estimation: Plot the function and zoom in near the point. The derivative at a point is the slope of the tangent line. Many tools allow you to:
- Draw or show the tangent line at a specific point.
- Use a built-in derivative tool to display \( f'(a) \) at selected values.
Example using a calculator: Estimate \( f'(2) \) for \( f(x) = \ln(x^2 + 1) \) using a calculator’s nDeriv feature:
On a TI calculator, enter: nDeriv(ln(x² + 1), x, 2) → Output: approx. 0.8
Benefit: Technology helps when the function is too complex to differentiate by hand or when only numerical/graphical data is available.
Example: Estimate \( f'(1) \) for \( f(x) = \sin(x^2) \) using a calculator or graphing software.
▶️Answer/Explanation
Using a TI calculator or online tool (e.g. Desmos or WolframAlpha):
\( f'(1) = \frac{d}{dx}[\sin(x^2)] \bigg|_{x=1} \)
Use numerical derivative: nDeriv(sin(x²), x, 1) → Output: approx. \( 1.0806 \)
Therefore, \( f'(1) \approx 1.081 \)