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Numerical differentiation

You should already be familiar with the idea of analytical differentiation and be able to differentiate simple functions like $y = x^{n}$ and $y = s i n (x)$ . If you don’t know how to do this, look it up now in any A level textbook or the relevant section of the Calculus Wikibook.

Sometimes functions are highly non-linear and a closed form for the derivative may be difficult to calculate. For example

$y = x^{\ln x}$

Alternatively, $y$ can be defined as the solution to an equation, so we can not calculate a closed form for $y = f (x)$ to which the traditional rules of differentiation may be applied. In such cases we may calculate a numerical approximation for the derivative using the following difference formulae.

Summary of mathematics used

Suppose $y = f (x)$ . Let the points $x_{0}, x_{1}, x_{2}, \dots, x_{N}$ be equally spaced over the interval $[a, b]$ , and let $h = \frac{1}{N} (x_{N} - x_{0}) = x_{i + 1} - x_{i}$ . Now let $y_{i} = f (x_{i})$ .

The forward difference approximation to the first derivative at $x = x_{i}$ is given by: $\frac{y_{i + 1} - y_{i}}{h}$
The backward difference approximation to the first derivative at $x = x_{i}$ is given by: $\frac{y_{i} - y_{i - 1}}{h}$
The central difference approximation to the first derivative at $x = x_{i}$ is given by: $\frac{y_{i + 1} - y_{i - 1}}{2 h}$
An approximation to the second derivative at $x = x_{i}$ is given by: $\frac{y_{i + 1} - 2 y_{i} + y_{i - 1}}{h^{2}}$

Where $h$ is assumed to be small, the smaller $h$ is, the better the approximation becomes. The derivation of these forms is undertaken in the next exercise.