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Find and prove a formula for the derivative of a product of n differentiable functions

Let

    \[ g = f_1 f_2 \cdots f_n \]

be the product of n functions f_1, \ldots, f_n with derivatives f'_1, \ldots, f'_n. Find a formula for the derivative of g, and prove that it is correct by mathematical induction.

Furthermore, show that

    \[ \frac{g'(x)}{g(x)} = \frac{f'_1 (x)}{f_1(x)} + \cdots + \frac{f'_n(x)}{f_n (x)}, \]

for those x at which f'_i(x) \neq 0 for any i = 1, \ldots, n.


Claim: If g = f_1 \cdots f_n, then

    \[g' = f'_1 (f_2 f_3 \cdots f_n) + f'_2 (f_1 f_3 \cdots f_n) + \cdots + f'_n(f_1 f_2 \cdots f_{n-1}). \]

Proof. For the case n = 2, from the usual product rule we have

    \[ g = f_1 f_2 \quad \implies \quad g' = f'_1 (f_2) + f'_2(f_1). \]

Thus, our claimed formula holds in this case. Assume then that it is true for some integer k \geq 2. Then, if g = f_1 \cdots f_{k+1} we have,

    \begin{align*}  &&g &= (f_1 \cdots f_k) f_{k+1}  \\  \implies && g' &= (f_1 \cdots f_k)' f_{k+1} + f'_{k+1} (f_1 \cdots f_k) & (\text{product rule}) \\  \implies && g' &= \left(f'_1 (f_2 \cdots f_k) + f'_2 (f_1 f_3 \cdots f_k) + \\  &&& \left. \qquad \cdots + f'_k (f_1 \cdots f_{k-1})\right) f_{k+1} + f'_{k+1} (f_1 \cdots f_k) & (\text{Ind. Hyp.})\\  \implies && g' &= f'_1 (f_2 \cdots f_{k+1}) + f'_2 (f_1 f_3 \cdots f_{k+1}) + \cdots + f'_{k+1}(f_1 \cdots f_k). \end{align*}

Therefore, if the formula holds for k then it also holds for k+1; thus, it holds for all positive integers. \qquad \blacksquare

Next, we prove the formula for the quotient, \frac{g'(x)}{g(x)}.

Proof. Let g be defined as above, and let x be a point at which f_i(x) \neq 0 for any i = 1, \ldots, n. Then,

    \begin{align*}  \frac{g'(x)}{g(x)} &= \frac{f'_1(f_2 \cdots f_n) + \cdots + f'_n(f_1 \cdots f_{n-1})}{f_1 \cdots f_n} \\  &= \frac{f'_1 \cdots f_n}{f_1 \cdots f_n} + \cdots + \frac{f_1 \cdots f'_n}{f_1 \cdots f_n} \\  &= \frac{f'_1}{f_1} + \cdots + \frac{f'_n}{f_n}. \qquad \blacksquare \end{align*}

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