Difference Formulas

The slope of the tangent line at $x$ can be approximated by the slope of a secant line through two nearby points. Depending on which pair of nearby points we choose, we get three distinct finite difference formulas. All three use a small step size $h$: the smaller $h$ is, the better the approximation (up to a limit imposed by rounding error, covered in a later section).

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Forward Difference

The forward difference formula uses the current point $x$ and the next point $x + h$:

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

It approximates the tangent slope by looking ahead by a step of size $h$. Geometrically, this draws a secant line connecting $f(x)$ and $f(x+h)$. As $h$ shrinks, the green secant line rotates downward to match the orange tangent line.

Worked Example

Problem: Using forward difference with $h = 0.1$, approximate $f'(2)$ for $f(x) = x^3 - 4x + 1$.

Step 1 — Identify the evaluation points

$x = 2, \qquad x + h = 2 + 0.1 = 2.1$

Step 2 — Evaluate $f$ at both points

$f(2) = (2)^3 - 4(2) + 1 = 8 - 8 + 1 = 1$

$f(2.1) = (2.1)^3 - 4(2.1) + 1 = 9.261 - 8.4 + 1 = 1.861$

Step 3 — Apply the formula

$f'(2) \approx \frac{f(2.1) - f(2)}{0.1} = \frac{1.861 - 1}{0.1} = \frac{0.861}{0.1} = \boxed{8.61}$

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Backward Difference

The backward difference formula uses the current point $x$ and the previous point $x - h$:

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

It approximates the tangent slope by looking behind by a step of size $h$. Geometrically, the secant line is drawn between the points $f(x-h)$ and $f(x)$.

Worked Example

Problem: Using backward difference with $h = 0.1$, approximate $f'(2)$ for $f(x) = x^3 - 4x + 1$.

Step 1 — Identify the evaluation points

$x = 2, \qquad x - h = 2 - 0.1 = 1.9$

Step 2 — Evaluate $f$ at both points

$f(2) = 1 \quad \text{(computed above)}$

$f(1.9) = (1.9)^3 - 4(1.9) + 1 = 6.859 - 7.6 + 1 = 0.259$

Step 3 — Apply the formula

$f'(2) \approx \frac{f(2) - f(1.9)}{0.1} = \frac{1 - 0.259}{0.1} = \frac{0.741}{0.1} = \boxed{7.41}$

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Central Difference

The central difference formula uses one point ahead of $x$ and one point behind $x$:

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

Because it is symmetric around $x$, the secant line spans from $f(x-h)$ all the way to $f(x+h)$. Visually, this resulting green secant line runs nearly parallel to the true orange tangent line, bypassing the localized curvature at $x$.

This is why the first-order errors from both sides partially cancel each other, making central difference significantly more accurate than forward or backward difference for the same step size $h$.

Worked Example

Problem: Using central difference with $h = 0.1$, approximate $f'(2)$ for $f(x) = x^3 - 4x + 1$.

Step 1 — Identify the evaluation points

$x + h = 2.1, \qquad x - h = 1.9$

Step 2 — Evaluate $f$ at both points

$f(2.1) = 1.861, \qquad f(1.9) = 0.259 \quad \text{(computed above)}$

Step 3 — Apply the formula

$f'(2) \approx \frac{f(2.1) - f(1.9)}{2(0.1)} = \frac{1.861 - 0.259}{0.2} = \frac{1.602}{0.2} = \boxed{8.01}$

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Comparison

The exact derivative at $x = 2$ is $f'(2) = 8$. All three methods use the same function and the same $h = 0.1$:

Method	Formula	Result	Error
Exact	$f'(2) = 3(2)^2 - 4$	$8.00$	$0.00$
Forward Difference	$\dfrac{f(x+h)-f(x)}{h}$	$8.61$	$0.61$
Backward Difference	$\dfrac{f(x)-f(x-h)}{h}$	$7.41$	$0.59$
Central Difference	$\dfrac{f(x+h)-f(x-h)}{2h}$	$8.01$	$0.01$

Note

Central difference produces the most accurate result by a large margin. For the same step size $h$, its error is proportional to $h^2$ rather than $h$, which is why it performs so much better. The next section proves this rigorously and quantifies the exact error bound for each method.