\begin{align} f'(x)=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}} \\ f''(x)=\lim_{h\to 0}{\frac{f(x+2h)-2f(x+h)+f(x)}{h^{2}}} \\ f'''(x)=\lim_{h\to 0}{\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}} \end{align}

And we are quick to see in general:

(2)\begin{align} f^{(n)}(x)=\lim_{h\to 0}{\frac{\sum_{i=0}^{n}{{n\choose i}(-1)^{n-i}f(x+ih)}}{h^{n}}} \end{align}

And that is pretty easy to put into BASIC code. Here is an issue, though. We want h→0, but we are dividing by h^{n}. Since the calcs loose accuracy after 14 digits, if we want to find the fifth derivative, we have to use an "h" value that is less than the fifth root of 1E-14, so our H value is not all that small D: I hope this helps!

**EDIT:** All you need is to use sum(seq()), though, to get that sum. If you have the newer OS, though, there is a Σ() command.

\begin{align} \operatorname{nDeriv}(f(t),t,x,h)=\frac{f(x+h)-f(x-h)}{2h} \end{align}

and nest *that*. I expect it to be an unholy mess, but if I was afraid of that, I wouldn't be using TI-Basic, would I?