Product of $n$ differentiable functions is differentiable

Question

1. Show for functions $f_1,\dots,f_n:A\rightarrow\mathbb{C}$ differentiable at $x_0\in A$ that $p(x)=\prod_{k=1}^{n}f_k(x)$ is differentiable at $x_0$ with $p'(x_0)=\sum_{j=1}^{n}f_j'(x_0)\prod_{k\neq j\land k=1}^{n}f_k(x_0)$.

2. Conclude that $\frac{p'(x_0)}{p(x_0)}=\sum_{j=1}^{n}\frac{f_j'(x_0)}{f_j(x_0)}$ if $p(x_0)\neq0$.

Answer 1

Question 1:

First, show it for just two functions.

Suppose that $f,g$ are differentiable at $x_0$. Then the following limits exist:

$$f'(x_0):=\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h} \; \; \; \; \; \; g'(x_0):=\lim_{h\rightarrow 0}\frac{g(x_0+h)-g(x_0)}{h}$$

Thus we compute the derivative of the product:

\begin{align} \ \bigl(fg\bigr)'(x_0)& = \lim_{h \rightarrow 0}\frac{f(x_0+h)g(x_0+h)-f(x_0)g(x_0)}{h} \\ \ & = \lim_{h \rightarrow 0}\frac{f(x_0+h)[g(x_0+h)-g(x_0)]+g(x_0)[f(x_0+h)-f(x_0)]}{h} \\ \ & = \lim_{h \rightarrow 0}f(x_0+h)\frac{g(x_0+h)-g(x_0)}{h}+\lim_{h \rightarrow 0}g(x_0)\frac{f(x_0+h)-f(x_0)}{h} \\ \ & = f(x_0)g'(x_0)+g(x_0)f'(x_0) \end{align}

which is the familiar "product rule" that we know.

What we have just proven is that

Theorem 1: If $f$ and $g$ are differentiable at $x_0$ with derivative $f'(x_0)$ and $g'(x_0)$, then their product is also differentiable with derivative $f(x_0)g'(x_0)+g(x_0)f'(x_0)$.

We then use Theorem 1 along with mathematical induction (I am only going to do the induction step here):

$\prod_{i=1}^kf_i \; \text{and} \; f_{k+1} \; \text{ differentiable at} \; x_0 \; \text{with derivative} \; \sum_{i=1}^kf_i'(x_0) \prod_{j=1,j \neq i}^kf_j(x_0) \; \text{and} \; f_{k+1}'(x_0) $

\begin{align} \ \implies \biggl(\prod_{i=1}^{k+1}f_i \biggr)'(x_0) & = \biggl(f_{k+1}\prod_{i=1}^{k}f_i \biggr)'(x_0)\\ \ & = f_{k+1}(x_0)\biggl(\prod_{i=1}^{k}f_i \biggr)'(x_0)+f_{k+1}'(x_0)\prod_{i=1}^{k}f_i(x_0) &&\text{by Theorem 1} \\ \ & = f_{k+1}(x_0)\sum_{i=1}^kf_i'(x_0) \prod_{j=1,j \neq i}^kf_j(x_0)+f_{k+1}'(x_0)\prod_{i=1}^{k}f_i (x_0) \\ \ & = \sum_{i=1}^kf_i'(x_0) \prod_{j=1,j \neq i}^{k+1}f_j(x_0)+f_{k+1}'(x_0)\prod_{i=1}^{k}f_i (x_0) \\ \ & = \sum_{i=1}^{k+1}f_i'(x_0) \prod_{j=1,j \neq i}^{k+1}f_j(x_0) \end{align}

as required.

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Question 2:

Direct computation suffices

\begin{align} \ \frac{p'(x_0)}{p(x_0)} & =\frac{\sum_{i=1}^{n}f_i'(x_0) \prod_{j=1,j \neq i}^{n}f_j(x_0)}{\prod_{j=1}^{n}f_j(x_0)} \\ \ & = \sum_{i=1}^{n}f_i'(x_0)\frac{\prod_{j=1,j \neq i}^{n}f_j(x_0)}{\prod_{j=1}^{n}f_j(x_0)} \\ \ & = \sum_{i=1}^{n}f_i'(x_0) \cdot \frac{1}{f_i(x_0)} \\ \ & = \sum_{i=1}^{n}\frac{f_i'(x_0)}{f_i(x_0)} \end{align}