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How do I show through mathematical induction that

\begin{equation} \frac{d^n}{dx^n}[f(x)\cdot g(x)] = \sum_{k = 0}^{n} \binom{n}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \end{equation}

which is the product rule for higher derivatives?

Chill2Macht
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Mestica
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    Have you proven things by induction before, or is that the major hurdle? – Matthew Leingang Jun 14 '16 at 14:03
  • @MatthewLeingang I know how to do mathematical induction, I was just intimidated with the equation here. – Mestica Jun 14 '16 at 19:06
  • In that case, it's relatively straightforward. The base case is clear. For the inductive step, apply the regular product rule one order below. Look for an identity for binomial coefficients that gives you what you want. :-) – Matthew Leingang Jun 14 '16 at 19:31

1 Answers1

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I assume you already know the base case, which is the standard product rule:

\begin{equation} \frac{d}{dx}[f(x)\cdot g(x)] = \frac{d}{dx} [f(x)] \cdot g(x) + f(x) \cdot \frac{d}{dx}[g(x)] \end{equation}

If you don't, here are two links for the necessary prerequisites:

Product Rule: https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html
Proof by Induction: http://www.purplemath.com/modules/inductn.htm

So now we need to state the induction hypothesis:

\begin{equation} \frac{d^{n-1}}{dx^{n-1}}[f(x)\cdot g(x)] = \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)] \end{equation}

and then prove that the induction step

\begin{equation} \frac{d^n}{dx^n}[f(x)\cdot g(x)] = \sum_{k = 0}^{n} \binom{n}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \end{equation}

holds, i.e. is true.

Allow us to apply the standard product rule onto the induction hypothesis and see what we find:

\begin{equation} \frac{d}{dx} \left[\frac{d^{n-1}}{dx^{n-1}}[f(x)\cdot g(x)]\right] = \frac{d}{dx} \left[ \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)] \right] \\ = \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d}{dx} \left[ \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)] \right] \quad \text{by linearity of the derivative} \\ = \sum_{k = 0}^{n-1} \left[ \binom{n-1}{k} \left[ \frac{d^{k+1}}{dx^{k+1}} [f(x)] \cdot \frac{d^{n-(k+1)}}{dx^{n-(k+1)}} [g(x)] \right] + \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] \right] \end{equation}

the last equality following by the standard product rule.

We now split the last sum into two sums

\begin{equation} \sum_{k=0}^{n-1} \binom{n-1}{k} \left[ \frac{d^{k+1}}{dx^{k+1}} [f(x)] \cdot \frac{d^{n-(k+1)}}{dx^{n-(k+1)}} [g(x)] \right] + \sum_{k=0}^{n-1} \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] \end{equation}

Reindexing the left sum we get

\begin{equation} \sum_{k=1}^{n-1} \binom{n-1}{k-1} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right]+ \frac{d^n}{dx^n}f(x) \cdot g(x) + \sum_{k=1}^{n-1} \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] + f(x) \cdot \frac{d^n}{dx^n} g(x) \end{equation}

We now use the Pascal rule for binomial coefficients

$$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

and combine the sums again for $k=1,\dots,n-1$, the values of $k$ which the two have in common:

\begin{equation} \sum_{k=1}^{n-1} \binom{n}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] + \frac{d^n}{dx^n}f(x) \cdot g(x) + f(x) \cdot \frac{d^n}{dx^n} g(x) \end{equation}

This is equal to

\begin{equation} \sum_{k=0}^{n} \binom{n}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] \end{equation}

Which is what we wanted to show.

Chill2Macht
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