Sorry but your questions make no sense, it is just a list of homework for a student of a basic course in numerical analysis and CFD.

Any good textbook explains the concept of consistence, local truncation error, discretization error, spectral accuracy and so on.

a. Why Numerical Techniques are Used for Finding Derivatives?

Numerical techniques are used to find derivatives when the analytical form of a function is either unavailable, difficult to differentiate, or when the function is based on experimental or real-world data points. Numerical differentiation is essential because it provides an approximation of the derivative by using finite differences between values, rather than relying on algebraic methods. This is particularly useful when dealing with complex systems in engineering, physics, or finance, where functions may not have a simple closed-form expression.

Numerical Differentiation is the process of estimating the derivative of a function using discrete data points. In numerical analysis and computation, it is applied to:

• Compute the rates of change in data sets (e.g., velocity from position data).
• Solve differential equations where derivatives are needed but not explicitly known.
• Perform sensitivity analysis in simulations, engineering models, and optimization.

b. Analytical vs. Numerical Definition of Derivatives

The analytical definition of a derivative for a function f(x)f(x)f(x) is the limit of the difference quotient as the interval hhh approaches zero:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}f′(x)=h→0lim​hf(x+h)−f(x)​

This definition assumes the function is continuous and differentiable in the interval of interest.

In contrast, numerical differentiation approximates the derivative using a small, finite value of hhh rather than taking the limit as h→0h \to 0h→0. The derivative is estimated as:

f′(x)≈f(x+h)−f(x)hf'(x) \approx \frac{f(x+h) - f(x)}{h}f′(x)≈hf(x+h)−f(x)​

c. Forward Difference Approximation

The forward difference approximation estimates the derivative by using the function value at a point xxx and the value at a small step forward, x+hx + hx+h:

f′(x)≈f(x+h)−f(x)hf'(x) \approx \frac{f(x+h) - f(x)}{h}f′(x)≈hf(x+h)−f(x)​

This method approximates the slope of the function at xxx using a small forward step hhh.

d. Forward Difference Example: Approximate Derivative of f(x)=x2f(x) = x^2f(x)=x2

For f(x)=x2f(x) = x^2f(x)=x2, we approximate f′(2)f'(2)f′(2) using the forward difference formula with different step sizes hhh.

f′(2)≈f(2+0.1)−f(2)0.1=(2.1)2−220.1=4.41−40.1=0.410.1=4.1f'(2) \approx \frac{f(2+0.1) - f(2)}{0.1} = \frac{(2.1)^2 - 2^2}{0.1} = \frac{4.41 - 4}{0.1} = \frac{0.41}{0.1} = 4.1f′(2)≈0.1f(2+0.1)−f(2)​=0.1(2.1)2−22​=0.14.41−4​=0.10.41​=4.1

f′(2)≈f(2+0.01)−f(2)0.01=(2.01)2−220.01=4.0401−40.01=0.04010.01=4.01f'(2) \approx \frac{f(2+0.01) - f(2)}{0.01} = \frac{(2.01)^2 - 2^2}{0.01} = \frac{4.0401 - 4}{0.01} = \frac{0.0401}{0.01} = 4.01f′(2)≈0.01f(2+0.01)−f(2)​=0.01(2.01)2−22​=0.014.0401−4​=0.010.0401​=4.01

Exact value: The derivative of f(x)=x2f(x) = x^2f(x)=x2 is f′(x)=2xf'(x) = 2xf′(x)=2x. So, f′(2)=4f'(2) = 4f′(2)=4.

Computational Errors:

• For h=0.1h = 0.1h=0.1: 4.1−4=0.14.1 - 4 = 0.14.1−4=0.1
• For h=0.01h = 0.01h=0.01: 4.01−4=0.014.01 - 4 = 0.014.01−4=0.01

e. Backward Difference Approximation

The backward difference approximation estimates the derivative using the function value at xxx and a small step backward, x−hx - hx−h:

f′(x)≈f(x)−f(x−h)hf'(x) \approx \frac{f(x) - f(x-h)}{h}f′(x)≈hf(x)−f(x−h)​

f. Backward Difference Example: Approximate Derivative of f(x)=x2f(x) = x^2f(x)=x2

f′(2)≈f(2)−f(2−0.1)0.1=4−(1.9)20.1=4−3.610.1=0.390.1=3.9f'(2) \approx \frac{f(2) - f(2-0.1)}{0.1} = \frac{4 - (1.9)^2}{0.1} = \frac{4 - 3.61}{0.1} = \frac{0.39}{0.1} = 3.9f′(2)≈0.1f(2)−f(2−0.1)​=0.14−(1.9)2​=0.14−3.61​=0.10.39​=3.9

f′(2)≈f(2)−f(2−0.01)0.01=4−(1.99)20.01=4−3.96010.01=0.03990.01=3.99f'(2) \approx \frac{f(2) - f(2-0.01)}{0.01} = \frac{4 - (1.99)^2}{0.01} = \frac{4 - 3.9601}{0.01} = \frac{0.0399}{0.01} = 3.99f′(2)≈0.01f(2)−f(2−0.01)​=0.014−(1.99)2​=0.014−3.9601​=0.010.0399​=3.99

Computational Errors:

• For h=0.1h = 0.1h=0.1: 4−3.9=0.14 - 3.9 = 0.14−3.9=0.1
• For h=0.01h = 0.01h=0.01: 4−3.99=0.014 - 3.99 = 0.014−3.99=0.01

g. Central Difference Approximation

The central difference approximation provides a more accurate estimate by using both forward and backward steps:

f′(x)≈f(x+h)−f(x−h)2hf'(x) \approx \frac{f(x+h) - f(x-h)}{2h}f′(x)≈2hf(x+h)−f(x−h)​

This method averages the forward and backward difference approximations, often resulting in a smaller error.

h. Central Difference Example: Approximate Derivative of f(x)=x2f(x) = x^2f(x)=x2

f′(2)≈f(2+0.1)−f(2−0.1)2(0.1)=4.41−3.610.2=0.80.2=4f'(2) \approx \frac{f(2+0.1) - f(2-0.1)}{2(0.1)} = \frac{4.41 - 3.61}{0.2} = \frac{0.8}{0.2} = 4f′(2)≈2(0.1)f(2+0.1)−f(2−0.1)​=0.24.41−3.61​=0.20.8​=4

f′(2)≈f(2+0.01)−f(2−0.01)2(0.01)=4.0401−3.96010.02=0.080.02=4f'(2) \approx \frac{f(2+0.01) - f(2-0.01)}{2(0.01)} = \frac{4.0401 - 3.9601}{0.02} = \frac{0.08}{0.02} = 4f′(2)≈2(0.01)f(2+0.01)−f(2−0.01)​=0.024.0401−3.9601​=0.020.08​=4

Computational Error: 4−4=04 - 4 = 04−4=0

i. Observation: Forward/Backward vs. Central Difference Approximation

• The forward and backward difference approximations introduce errors depending on the direction and step size.
• The central difference approximation is generally more accurate because it balances errors by using both forward and backward differences.
• Smaller step sizes reduce errors, but the central difference approximation still outperforms forward and backward differences for the same step size.

j. Second Derivative Approximation

The second derivative approximation using numerical differentiation is often computed using the central difference formula:

f′′(x)≈f(x+h)−2f(x)+f(x−h)h2f''(x) \approx \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}f′′(x)≈h2f(x+h)−2f(x)+f(x−h)​

This estimates the curvature or concavity of the function at xxx.

For f(x)=x2f(x) = x^2f(x)=x2, f′′(x)=2f''(x) = 2f′′(x)=2, and applying the second derivative approximation for f(2)f(2)f(2) with h=0.1h = 0.1h=0.1:

f′′(2)≈(2.1)2−2(2)2+(1.9)2(0.1)2=4.41−8+3.610.01=0.020.01=2f''(2) \approx \frac{(2.1)^2 - 2(2)^2 + (1.9)^2}{(0.1)^2} = \frac{4.41 - 8 + 3.61}{0.01} = \frac{0.02}{0.01} = 2f′′(2)≈(0.1)2(2.1)2−2(2)2+(1.9)2​=0.014.41−8+3.61​=0.010.02​=2

This matches the exact value f′′(2)=2f''(2) = 2f′′(2)=2, showing no computational error.