I know you can get the derivative of a function with nDeriv(, but for some reason, as the page notes, you can only nest nDeriv( once. This means that while it is easy to find *f*'(x) and *f*"(x), there is no obvious way to find *f*^{3}(x) or *f*^{4}(x). This seems like an incredible oversight on TI's part, and I was wondering if there was a workaround. Unfortunately I can't simply find the second derivative of the function at a point, store the output in a variable, and find the second derivative of that, because I'm trying to use it within an fMax( function.

As I see that you're a member from a long time ago, I must say welcome back :) Unfortunately, that's about where I must end since I unfortunately know very little about the nDeriv( command :/ there are people that do know though, and as soon as one of them happens to drop by, they'll be able to help you out :D

Okay, I think I may have come up with a solution. Instead of using the nDeriv( function four times, I'm simply going to expand it to the formula used to calculate it

(1)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?

I was thinking that, too, but it is even more of a mess than that! If you want the second derivative, you have to find 3 close points and find the slopes of each to create a quadratic equation as a good estimate for the derivative on that interval, then you need to find the derivative at the point on your quadratic! Now imagine if you wanted to find the third derivative. This will become recursive very quickly and eat up tons of memory o.o It would actually be better to program a symbolic derivative calculator, even though that may seem tedious. You could also use the Symbolic app which comes with a derivative function that finds nth derivatives.

Z80 Assembly>English>TI-BASIC>Python>French>C>0

Psh, I must have been really tired this morning to think of that ^ I actually made a program to compute that mess, too, until I realised how much I overcomplicated it XD After some substitution, we have:

(1)And we are quick to see in general:

(2)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.

Z80 Assembly>English>TI-BASIC>Python>French>C>0