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,

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

One comment

September 22, 2021 at 12:52 pm

Gonzalo Spina says:

Well I didn’t arrive to the same formula. I’ve arrived into a recursive formula which is:

$\\G_n(x) = G^I_{n-1}(x).F_n(x) + G_{n-1}(x).F^I_n(x)$

And can also be proved by induction

Stumbling Robot

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

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