Numerical differentiation in MATLAB - Step-by-Step Execution

Concept Flow - Numerical differentiation

Start with function f(x)

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Choose point x0 and step h

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Calculate f(x0 + h) and f(x0 - h)

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Apply formula: (f(x0 + h) - f(x0 - h)) / (2*h)

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Result: Approximate derivative at x0

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End

Numerical differentiation approximates the slope of a function at a point by using values of the function near that point.

Execution Sample

MATLAB

f = @(x) x.^2;
x0 = 2;
h = 0.01;
derivative = (f(x0 + h) - f(x0 - h)) / (2*h);

This code approximates the derivative of f(x) = x^2 at x = 2 using a small step h.

Execution Table

Step	Expression	Evaluation	Result
1	Calculate f(x0 + h)	f(2 + 0.01)	4.0401
2	Calculate f(x0 - h)	f(2 - 0.01)	3.9601
3	Subtract f(x0 + h) - f(x0 - h)	4.0401 - 3.9601	0.08
4	Divide by 2*h	0.08 / (2*0.01)	4
5	Result	Derivative at x0	4

💡 Finished calculation: derivative approximated as 4 at x = 2

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
f(x0 + h)	-	4.0401	4.0401	4.0401	4.0401	4.0401
f(x0 - h)	-	-	3.9601	3.9601	3.9601	3.9601
Numerator	-	-	-	0.08	0.08	0.08
Denominator	-	-	-	-	0.02	0.02
derivative	-	-	-	-	-	4

Key Moments - 2 Insights

Why do we use (f(x0 + h) - f(x0 - h)) instead of just (f(x0 + h) - f(x0))?

Using values on both sides (x0 + h and x0 - h) gives a better approximation by balancing errors, as shown in steps 1 and 2 of the execution_table.

What happens if h is too large or too small?

If h is too large, the approximation is less accurate; if too small, rounding errors increase. The chosen h=0.01 balances this, as seen in the stable results in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of f(x0 - h) at step 2?

A 0.08

B 4.0401

C 3.9601

D 4

💡 Hint

Check the 'Evaluation' column in row for step 2 in execution_table.

At which step is the numerator (f(x0 + h) - f(x0 - h)) calculated?

A Step 1

B Step 3

C Step 4

D Step 5

💡 Hint

Look for the subtraction operation in the 'Expression' column in execution_table.

If h was doubled to 0.02, how would the denominator in step 4 change?

A It would become 0.04

B It would become 0.01

C It would stay 0.02

D It would become 0.08

💡 Hint

Denominator is 2*h, so doubling h doubles the denominator (see step 4 denominator in variable_tracker).

Concept Snapshot

Numerical differentiation approximates f'(x0) by (f(x0 + h) - f(x0 - h)) / (2*h).
Choose a small step h for accuracy.
This is called the central difference method.
It uses function values near x0 to estimate slope.
Useful when derivative formula is unknown or complex.

Full Transcript

Numerical differentiation finds the slope of a function at a point by using nearby function values. We pick a point x0 and a small step h. Then we calculate the function at x0 + h and x0 - h. Subtract these two values and divide by 2*h. This gives an approximate derivative at x0. For example, for f(x) = x^2 at x=2 with h=0.01, the derivative is about 4. This method balances errors by using points on both sides. Choosing h carefully is important to avoid big errors or rounding problems.