Question

Let \( f(x) = \ln x \). The first derivative \( f'(x) \) is to be calculated at \( x = 1 \) using numerical differentiation. \( f'(1) \) is calculated using first order forward difference (\( f'_{FD} \)), first order backward difference (\( f'_{BD} \)), and second order central difference (\( f'_{CD} \)), using interval width \( h = 0.1 \). The CORRECT order of the values of \( f'_{FD} \), \( f'_{BD} \), and \( f'_{CD} \) is:

• A. \( f'_{BD}>f'_{CD}>f'_{FD} \)
• B. \( f'_{CD}>f'_{BD}>f'_{FD} \)
• C. \( f'_{BD}>f'_{FD}>f'_{CD} \)
• D. \( f'_{BD}>f'_{FD}>f'_{CD} \)

Hint:

In numerical differentiation, first order backward differences tend to provide more accurate estimates than forward differences, and second order central differences are often more accurate than both.

Answer 1

The Correct Option is D

We are tasked with calculating the first derivative of the function \( f(x) = \ln x \) at \( x = 1 \) using numerical differentiation methods. The exact value of the first derivative of \( f(x) = \ln x \) is: \[ f'(x) = \frac{1}{x} \] At \( x = 1 \), the exact value of \( f'(1) \) is: \[ f'(1) = \frac{1}{1} = 1 \] Now, let's calculate the approximate values of \( f'(1) \) using the three methods mentioned: 1. First Order Forward Difference (\( f'_{FD} \)): The first order forward difference formula is given by: \[ f'_{FD} = \frac{f(x+h) - f(x)}{h} \] Substitute \( x = 1 \) and \( h = 0.1 \): \[ f'_{FD} = \frac{f(1+0.1) - f(1)}{0.1} = \frac{\ln(1.1) - \ln(1)}{0.1} = \frac{0.095310 - 0}{0.1} = 0.9531 \] 2. First Order Backward Difference (\( f'_{BD} \)): The first order backward difference formula is given by: \[ f'_{BD} = \frac{f(x) - f(x-h)}{h} \] Substitute \( x = 1 \) and \( h = 0.1 \): \[ f'_{BD} = \frac{f(1) - f(1-0.1)}{0.1} = \frac{\ln(1) - \ln(0.9)}{0.1} = \frac{0 - (-0.105360)}{0.1} = 1.0536 \] 3. Second Order Central Difference (\( f'_{CD} \)): The second order central difference formula is given by: \[ f'_{CD} = \frac{f(x+h) - f(x-h)}{2h} \] Substitute \( x = 1 \) and \( h = 0.1 \): \[ f'_{CD} = \frac{f(1+0.1) - f(1-0.1)}{2 \times 0.1} = \frac{\ln(1.1) - \ln(0.9)}{0.2} = \frac{0.095310 - (-0.105360)}{0.2} = 1.0023 \] Comparing the Results: - The exact value of \( f'(1) \) is 1. - \( f'_{FD} = 0.9531 \) - \( f'_{BD} = 1.0536 \) - \( f'_{CD} = 1.0023 \) Thus, the order of the approximations is: \[ f'_{BD}>f'_{FD}>f'_{CD} \] The correct order is \( \boxed{D} \).