Quote:
Originally Posted by Slavakion
Is there a shortcut to finding second/third/... derivatives? Not that it's difficult (yet), but it gets tedious. Especially when you get the evil ones with multiple chain rules.
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Well, you can memorize certain formulas. For example, in the case of the nth derivative of x^k, where n is a nonnegative integer, we have
D^n x^k = (k)_n x^{k-n}, where
(k)_n is the falling factorial of k to n, namely,
(k)_n = k (k-1) (k-2) ... (k-n+1) if n>0;
(k)_0 = 1.
Another useful formula is Leibniz' Rule, which is a generalization of the product rule, and looks like this:
D^n (f(x)g(x)) = \sum_{i=0}^n \binom{n}{i} f^(i)(x) g^(n-i)(x),
where \binom{n}{i} is the binomial coefficient, and f^(i) is the ith derivative of f. Note its relation to the binomial theorem, making it easy to remember. Graphically,
There's also a generalization of the chain rule, but I can't remember it and it's sort of ugly anyway.