Numerical differentiation in MATLAB - Mini Project: Build & Apply

Numerical differentiation

📖 Scenario: You are working as a data analyst and need to find the rate of change of a function at specific points using numerical methods. This is useful when you have data points but no explicit formula for the function.

🎯 Goal: Build a MATLAB program that calculates the numerical derivative of a function at given points using the forward difference method.

📋 What You'll Learn

Create a vector of x values

Create a vector of corresponding y values for the function y = x^2

Define the step size h

Calculate the numerical derivative using the forward difference formula

Print the resulting derivative vector

💡 Why This Matters

🌍 Real World

Numerical differentiation helps estimate rates of change when you only have data points, such as measuring speed from position data.

💼 Career

Data analysts, engineers, and scientists often use numerical differentiation to analyze experimental or simulation data where formulas are unknown.

1

Create the data vectors

Create a vector called x with values 1, 2, 3, 4, 5. Then create a vector called y where each element is the square of the corresponding x value.

MATLAB

% Your code here

Hint

Use the dot operator .^ to square each element of x individually.

2

Define the step size

Create a variable called h and set it to 1, which is the difference between consecutive x values.

MATLAB

x = [1, 2, 3, 4, 5];
y = x.^2;
% Your code here

Hint

The step size h is the difference between two consecutive x values.

3

Calculate the numerical derivative

Create a vector called dy_dx that stores the numerical derivative of y with respect to x using the forward difference formula: dy_dx(i) = (y(i+1) - y(i)) / h for i from 1 to 4.

MATLAB

x = [1, 2, 3, 4, 5];
y = x.^2;
h = 1;
% Your code here

Hint

Use a for loop from 1 to length of y minus 1. Calculate the difference quotient inside the loop.

4

Display the derivative

Use disp to print the vector dy_dx which contains the numerical derivatives.

MATLAB

x = [1, 2, 3, 4, 5];
y = x.^2;
h = 1;
dy_dx = zeros(1, length(y)-1);
for i = 1:length(y)-1
    dy_dx(i) = (y(i+1) - y(i)) / h;
end
% Your code here

Hint

Use disp(dy_dx) to show the derivative values.

Practice

1. Which method in numerical differentiation uses points on both sides of the target point to estimate the derivative?

easy

A. Forward difference method

B. Euler's method

C. Backward difference method

D. Central difference method

Solution

1. Step 1: Understand difference methods

   Forward difference uses the point and the next point; backward difference uses the point and the previous point.

2. Step 2: Identify method using both sides

   Central difference uses points before and after the target point for better accuracy.

3. Final Answer:

   Central difference method -> Option D

4. Quick Check:

   Both sides points = Central difference [OK]

Hint: Central difference uses points on both sides [OK]

Common Mistakes:

• Confusing forward with central difference
• Thinking Euler's method is for differentiation
• Assuming backward difference uses both sides

2. Which of the following MATLAB code snippets correctly computes the forward difference approximation of the derivative of function f at point x with step size h?

easy

A. df = (f(x+h) - f(x)) / h;

B. df = (f(x) - f(x-h)) / h;

C. df = (f(x+h) - f(x-h)) / (2*h);

D. df = (f(x) + f(x+h)) / h;

Solution

1. Step 1: Recall forward difference formula

   Forward difference formula is (f(x+h) - f(x)) / h.

2. Step 2: Match code to formula

   df = (f(x+h) - f(x)) / h; matches this formula exactly.

3. Final Answer:

   df = (f(x+h) - f(x)) / h; -> Option A

4. Quick Check:

   Forward difference = (f(x+h)-f(x))/h [OK]

Hint: Forward difference subtracts current from next point [OK]

Common Mistakes:

• Using backward difference formula instead
• Dividing by 2*h instead of h
• Adding function values instead of subtracting

3. Consider the MATLAB code:

f = @(x) x^2;
x = 2;
h = 0.01;
df = (f(x+h) - f(x-h)) / (2*h);

What is the value of df approximately?

medium

A. 5.00

B. 2.00

C. 4.00

D. 3.99

Solution

1. Step 1: Calculate f(x+h) and f(x-h)

   f(2+0.01) = (2.01)^2 = 4.0401; f(2-0.01) = (1.99)^2 = 3.9601.

2. Step 2: Compute central difference

   df = (4.0401 - 3.9601) / (2*0.01) = 0.08 / 0.02 = 4.00.

3. Step 3: MATLAB precision

   In MATLAB, df is computed and displayed as 4.0000.

4. Final Answer:

   4.00 -> Option C

5. Quick Check:

   Central difference ≈ 4 [OK]

Hint: Central difference approximates slope near x [OK]

Common Mistakes:

• Using forward difference formula instead
• Forgetting to divide by 2*h
• Incorrect function evaluation

4. The following MATLAB code attempts to compute the derivative using forward difference but gives incorrect results:

f = @(x) sin(x);
x = pi/4;
h = 0.001;
df = (f(x) - f(x+h)) / h;

What is the error in this code?

medium

A. Step size h is too large

B. The subtraction order is reversed; should be f(x+h) - f(x)

C. Function handle syntax is incorrect

D. Variable x should be a vector

Solution

1. Step 1: Recall forward difference formula

   Forward difference is (f(x+h) - f(x)) / h, not (f(x) - f(x+h)) / h.

2. Step 2: Identify error in code

   The code subtracts f(x+h) from f(x), reversing the order and causing incorrect sign.

3. Final Answer:

   The subtraction order is reversed; should be f(x+h) - f(x) -> Option B

4. Quick Check:

   Forward difference = (f(x+h)-f(x))/h [OK]

Hint: Forward difference subtracts future minus current value [OK]

Common Mistakes:

• Reversing subtraction order
• Using too large step size without reason
• Misunderstanding function handle syntax

5. You want to estimate the derivative of f(x) = exp(-x^2) at x = 1 using central difference in MATLAB. Which code snippet correctly implements this with step size h = 0.0001?

hard

A. f = @(x) exp(-x^2);
x = 1;
h = 0.0001;
df = (f(x+h) - f(x-h)) / (2*h);

B. f = @(x) exp(-x^2);
x = 1;
h = 0.0001;
df = (f(x) - f(x-h)) / h;

C. f = @(x) exp(-x^2);
x = 1;
h = 0.0001;
df = (f(x+h) + f(x-h)) / (2*h);

D. f = @(x) exp(-x^2);
x = 1;
h = 0.0001;
df = (f(x+h) - f(x)) / h;

Solution

1. Step 1: Recall central difference formula

   Central difference formula is (f(x+h) - f(x-h)) / (2*h) for better accuracy.

2. Step 2: Match code to formula

   f = @(x) exp(-x^2); x = 1; h = 0.0001; df = (f(x+h) - f(x-h)) / (2*h); matches the central difference formula exactly with correct function and step size.

3. Final Answer:

   f = @(x) exp(-x^2); x = 1; h = 0.0001; df = (f(x+h) - f(x-h)) / (2*h); -> Option A

4. Quick Check:

   Central difference = (f(x+h)-f(x-h))/(2*h) [OK]

Hint: Central difference uses symmetric points around x [OK]

Common Mistakes:

• Using forward or backward difference instead
• Adding function values instead of subtracting
• Dividing by h instead of 2*h