(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'm told to find the nth derivative of a function via Leibniz.

2. Relevant equations

Then I'm asked to show that f^{(n+1)}=f^{(n)}+3n(n-1)f^{(n-2)}evaluated at x=0 when n>1

3. The attempt at a solution

I can solve this just fine when plugging in a constant for n. For example, when n=2, f '" = f " + 6 f, no problem.My question though is whether this is enough to prove that the above equation balances for all n>1?

If not, how would I show this? Would I need to use the general form of f^{(n)}I found via the Leibniz formula? How would this work? I ask this because it is of the form f^{(n)}= Ʃ_{k=0}^{n}(n!/k!(n-k!) * (d^{k}/dx^{k}*f) * (d^{n-k}/dx^{n-k}*g)) and am not sure how I would transform that into f^{(n+1)}or f^{(n-2)}with the summation and combination.

Thank you in advance!

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# Homework Help: Finding the nth+1 derivative via Leibniz

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