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Prove Leibniz’s formula

If h(x) = f(x) g(x) is a product of functions prove that the derivatives of h(x) are given by the formula

    \[ h^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} f^{(k)} (x) g^{(n-k)}(x), \]


    \[ \binom{n}{k} = \frac{n!}{k! (n-k)!} \]

is the binomial coefficient. (See the first four exercises of Section I.4.10 on page 44 of Apostol, and in particular see Exercise #4, in which we prove the binomial theorem.)

Proof. The proof is by induction. Letting h(x) = f(x) g(x) and n = 1 we use the product rule for derivatives

    \[ h'(x) = f(x)g'(x) + f'(x)g(x) = \sum_{k=0}^1 \binom{1}{k} f^{(k)}(x) g^{(1-k)}(x). \]

So, the formula is true for the case n = 1. Assume then that it is true for n = m \in \mathbb{Z}_{> 0}. Then we consider the (m+1)st derivative h^{(m+1)}(x):

    \begin{align*}  h^{(m+1)}(x) &= \left( h^{(m)}(x) \right)' \\  &= \left( \sum_{k=0}^m \binom{m}{k} f^{(k)}(x) g^{(m-k)}(x) \right)' &(\text{Ind. Hyp.})\\  &= \sum_{k=0}^m \left( \binom{m}{k} f^{(k)}(x) g^{(m-k)}(x) \right)'. \end{align*}

Here, we use linearity of the derivative to differentiate term by term over this finite sum. This property was established in Theorem 4.1 (i) and the comments following the theorem on page 164 of Apostol. Continuing where we left off we apply the product rule,

    \begin{align*}  &= \sum_{k=0}^m \left( \binom{m}{k} \left(f^{(k+1)}(x) g^{(m-k)}(x) + f^{(k)}(x) g^{(m-k+1)}(x) \right) \right) \\[9pt]  &= \sum_{k=0}^m \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x) + \sum_{k=0}^m \binom{m}{k} f^{(k)}(x) g^{(m-k+1)}(x) \\[9pt]  &= \sum_{k=1}^{m+1} \binom{m}{k-1} f^{(k)} g^{(m-k+1)}(x) + \sum_{k=0}^m \binom{m}{k} f^{(k)}(x) g^{(m-k+1)}(x), \end{align*}

where we’ve reindexed the first sum to run from k = 1 to m + 1 instead of from k = 0 to m. Then, we pull out the k = m+1 term from the first sum and the k = 0 term from the second,

    \begin{align*}  &= f^{(m+1)}(x) g(x) + \sum_{k=1}^m \binom{m}{k-1} f^{(k)}(x) g^{(m-k+1)}(x) + \sum_{k=1}^m \binom{m}{k} f^{(k)}(x) g^{(m-k+1)}(x) + f(x) g^{(m+1)}(x) \\  &= f^{(m+1)}(x) g(x) + f(x)g^{(m+1)}(x) + \sum_{k=1}^m \left( \binom{m}{k-1} + \binom{m}{k} \right) f^{(k)}(x) g^{(m-k+1)} (x). \end{align*}

Now, we recall the law of Pascal’s triangle (which we proved in a previous exercise) which establishes that

    \[ \binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}. \]

Therefore, we have

    \begin{align*}  &= f^{(m+1)}(x) g(x) + f(x) g^{(m+1)}(x) + \sum_{k=1}^m  \binom{m+1}{k} f^{(k)}(x)g^{(m-k+1)}(x) \\  &= \sum_{k=0}^{m+1} \binom{m+1}{k} f^{(k)}(x) g^{(m-k+1)}(x). \end{align*}

Hence, the formula holds for m+1 if it holds m. Therefore, we have established that it holds for all positive integers n.

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