Numerical Differentiation (Central Difference Method)

Numerical Differentiation is a technique used to estimate the derivatives of a function $f(x)$ at a specific point $x$ using values of the function evaluated at neighboring points spaced by a step size $h$ .

The Central Difference Formulas achieve $\mathcal{O}(h^2)$ second-order accuracy by averaging forward and backward steps:

First Derivative Formula:

f'(x) \approx \frac{f(x + h) - f(x - h)}{2h}

Second Derivative Formula:

f''(x) \approx \frac{f(x + h) - 2f(x) + f(x - h)}{h^2}

Where:

$x$ is the target point at which derivatives are evaluated.
$h$ is the step size.
$f(x + h)$ and $f(x - h)$ are forward and backward evaluations.

Example of Numerical Differentiation

Given Function

f(x) = x^3 - 2x + 5

At $x = 2, h = 0.1$

Step 1: Evaluate Function at Key Points

f(1.9) = (1.9)^3 - 2(1.9) + 5 = 8.059

f(2.0) = (2.0)^3 - 2(2.0) + 5 = 9.000

f(2.1) = (2.1)^3 - 2(2.1) + 5 = 10.061

Step 2: First Derivative ( $f'(2)$ )

f'(2) \approx \frac{10.061 - 8.059}{0.2} = 10.0100

Step 3: Second Derivative ( $f''(2)$ )

f''(2) \approx \frac{10.061 - 2(9.0) + 8.059}{0.01} = 12.0000

Conclusion

\boxed{f'(2) \approx 10.0100, \quad f''(2) \approx 12.0000}

Therefore, using Central Difference Numerical Differentiation with $h = 0.1$ , the estimated derivatives are f'(2) = 10.0100 and f''(2) = 12.0000 (exact values 10 and 12).

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Numerical Differentiation (Central Difference Method)

The Central Difference Formulas achieve $\mathcal{O}(h^2)$ second-order accuracy by averaging forward and backward steps:

First Derivative Formula:

f'(x) \approx \frac{f(x + h) - f(x - h)}{2h}

Second Derivative Formula:

f''(x) \approx \frac{f(x + h) - 2f(x) + f(x - h)}{h^2}

Where:

$x$ is the target point at which derivatives are evaluated.
$h$ is the step size.
$f(x + h)$ and $f(x - h)$ are forward and backward evaluations.

Example of Numerical Differentiation

Given Function

f(x) = x^3 - 2x + 5

At $x = 2, h = 0.1$

Step 1: Evaluate Function at Key Points

f(1.9) = (1.9)^3 - 2(1.9) + 5 = 8.059

f(2.0) = (2.0)^3 - 2(2.0) + 5 = 9.000

f(2.1) = (2.1)^3 - 2(2.1) + 5 = 10.061

Step 2: First Derivative ( $f'(2)$ )

f'(2) \approx \frac{10.061 - 8.059}{0.2} = 10.0100

Step 3: Second Derivative ( $f''(2)$ )

f''(2) \approx \frac{10.061 - 2(9.0) + 8.059}{0.01} = 12.0000

Conclusion

\boxed{f'(2) \approx 10.0100, \quad f''(2) \approx 12.0000}

Therefore, using Central Difference Numerical Differentiation with $h = 0.1$ , the estimated derivatives are f'(2) = 10.0100 and f''(2) = 12.0000 (exact values 10 and 12).

Interactive Numerical Differentiation Calculator

Numerical Differentiation Solver

PreviousWeddle's Rule

Unit 3•17 of 28

NextTaylor's Series Method

Numerical Differentiation (Central Difference Method)

Example of Numerical Differentiation

Step 1: Evaluate Function at Key Points

Step 2: First Derivative (f′(2)f'(2)f′(2))

Step 3: Second Derivative (f′′(2)f''(2)f′′(2))

Conclusion

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Numerical Differentiation (Central Difference Method)

Example of Numerical Differentiation

Step 1: Evaluate Function at Key Points

Step 2: First Derivative (f′(2)f'(2)f′(2))

Step 3: Second Derivative (f′′(2)f''(2)f′′(2))

Conclusion

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Step 2: First Derivative ( $f'(2)$ )

Step 3: Second Derivative ( $f''(2)$ )

Step 2: First Derivative ( $f'(2)$ )

Step 3: Second Derivative ( $f''(2)$ )