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 (\(f'_{BD}\)), - Second-order central difference (\(f'_{CD}\)), with interval width \(h = 0.1\). The correct order of the values of \(f'_{FD}, f'_{BD}, f'_{CD}\) is:

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

Hint:

Forward difference underestimates, backward difference overestimates, while central difference gives the most accurate result (especially for small \(h\)).

Answer 1

The Correct Option is A

Step 1: True derivative at \(x=1\).
\[ f(x) = \ln x, \quad f'(x) = \frac{1}{x}, \quad f'(1) = 1 \] Step 2: Forward difference approximation.
\[ f'_{FD}(1) = \frac{f(1+h) - f(1)}{h} \] \[ = \frac{\ln(1.1) - \ln(1)}{0.1} = \frac{\ln(1.1)}{0.1} \] \[ = \frac{0.09531}{0.1} = 0.9531 \] Step 3: Backward difference approximation.
\[ f'_{BD}(1) = \frac{f(1) - f(1-h)}{h} \] \[ = \frac{\ln(1) - \ln(0.9)}{0.1} = \frac{0 - (-0.10536)}{0.1} \] \[ = \frac{0.10536}{0.1} = 1.0536 \] Step 4: Central difference approximation.
\[ f'_{CD}(1) = \frac{f(1+h) - f(1-h)}{2h} \] \[ = \frac{\ln(1.1) - \ln(0.9)}{0.2} \] \[ = \frac{0.09531 - (-0.10536)}{0.2} = \frac{0.20067}{0.2} = 1.00335 \] Step 5: Compare results.
- \(f'_{FD}(1) = 0.9531\) (underestimates). - \(f'_{CD}(1) = 1.0033\) (very close to true). - \(f'_{BD}(1) = 1.0536\) (overestimates). So, order is: \[ f'_{BD} > f'_{CD} > f'_{FD}. \] Final Answer: \[ \boxed{\text{(D) } f'_{BD} > f'_{CD} > f'_{FD}} \]