# The Nth Derivative of a Function (General Form)

by RocketSurgery
Tags: derivative, form, function
 HW Helper P: 2,566 That formula isn't quite applicable to this problem, since you can only take the factorial of integers. There is a generalization of the factorial to non-integers which makes that true for all k, but it's a little complicated, and not worth the effort for this problem (unless you're interested). Do you know the derivation of that formula? You can use induction to prove it, but it's simple to see where it comes from by just writing out the first few derivatives in general form: $$\frac{d}{dx}\left(x^k \right) = k x^{k-1}$$ $$\frac{d^2}{dx^2}\left(x^k \right) = k(k-1) x^{k-2}$$ ... $$\frac{d^a}{dx^a}\left(x^k \right) = k(k-1)...(k-a+1) x^{k-a}$$ This is true for any k. Then if k was an integer we'd notice that: $$k(k-1)...(k-a+1) = \frac{k(k-1)...(k-a+1)(k-a)...(2)(1)}{(k-a)...(2)(1)} = \frac{k!}{(k-a)!}$$ which gives the formula. When k is not an integer, you can't do this. But when k is 1/2, you can do something like it: $$\frac{1}{2}\left(\frac{1}{2}-1\right) = \frac{1}{2}\left(-\frac{1}{2}\right) = - \frac{1}{4}$$ $$\frac{1}{2}\left(\frac{1}{2}-1\right) \left(\frac{1}{2}-2\right) = \frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \frac{3}{8}$$ Can you see where this is going, and come up with a formula for k(k-1)...(k-a+1) when k=1/2?