Numerical differentiation in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of forward difference approximation

What is the output of this MATLAB code that approximates the derivative of f(x) = x^2 at x = 2 using the forward difference method with h = 0.01?

MATLAB

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

A 4.01

B 4.00

C 3.99

D Error: Undefined function or variable 'f'

💡 Hint

Recall the formula for forward difference: (f(x+h) - f(x)) / h.

The exact derivative of x^2 at x=2 is 4. The forward difference with h=0.01 gives approximately (4.0401 - 4)/0.01 = 4.01.

❓ Predict Output

intermediate

Output of central difference approximation

What is the output of this MATLAB code that approximates the derivative of f(x) = sin(x) at x = pi/4 using the central difference method with h = 0.001?

MATLAB

f = @(x) sin(x);
x = pi/4;
h = 0.001;
derivative = (f(x + h) - f(x - h)) / (2*h);
disp(derivative);

A 0.7060

B 0.7070

C 0.7071

D Error: Index exceeds matrix dimensions.

💡 Hint

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

The exact derivative of sin(x) at pi/4 is cos(pi/4) ≈ 0.7071. The central difference with h=0.001 approximates this closely.

🔧 Debug

advanced

Identify the error in backward difference code

This MATLAB code attempts to compute the derivative of f(x) = exp(x) at x = 1 using the backward difference method with h = 0.01. What error will it produce when run?

MATLAB

f = @(x) exp(x);
x = 1;
h = 0.01;
derivative = (f(x) - f(x - h)) / h;

A Index exceeds matrix dimensions.

B No error; outputs approximately 2.7183

C SyntaxError: Missing semicolon

D Undefined function or variable 'h'

💡 Hint

Check if the code syntax is correct and variables are defined.

The code correctly computes the backward difference and outputs approximately exp(1) ≈ 2.7183. No error occurs.

❓ Predict Output

advanced

Output of second derivative approximation

What is the output of this MATLAB code that approximates the second derivative of f(x) = x^3 at x = 2 using the central difference formula with h = 0.001?

MATLAB

f = @(x) x.^3;
x = 2;
h = 0.001;
second_derivative = (f(x + h) - 2*f(x) + f(x - h)) / (h^2);
disp(second_derivative);

A 12.0000

B Error: Undefined function or variable 'second_derivative'

C 13.0000

D 11.9990

💡 Hint

Recall the second derivative central difference formula: (f(x+h) - 2f(x) + f(x-h)) / h^2.

The exact second derivative of x^3 is 6x. At x=2, it is 12. The central difference with h=0.001 gives approximately 12.0000.

🧠 Conceptual

expert

Error behavior in numerical differentiation

Which statement correctly describes the error behavior when using the central difference method to approximate the first derivative of a smooth function as the step size h approaches zero?

A The error increases exponentially as h decreases.

B The error is constant regardless of h.

C The error always decreases as h approaches zero without limit.

D The error decreases linearly with h and then increases due to round-off errors for very small h.

💡 Hint

Consider both truncation and round-off errors in numerical differentiation.

Central difference error has two parts: truncation error decreases with smaller h, but round-off error increases for very small h due to floating-point precision limits, causing total error to first decrease then increase.