# Product Rule for Derivatives/General Result

## Theorem

Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions differentiable on the open interval $I$.

then:

$\forall x \in I: \ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$

### $3$ Factors

Let $\map u x$, $\map v x$ and $\map w x$ be real functions differentiable on the open interval $I$.

Then:

$\forall x \in I: \map {\dfrac \d {\d x} } {u v w} = u v \dfrac {\d w} {\d x} + u w \dfrac {\d v} {\d x} + v w \dfrac {\d u} {\d x}$

## Proof

Proof by Principle of Mathematical Induction:

For all $n \in \N_{\ge 1}$, let $\map P n$ be the proposition:

$\ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$

$\map P 1$ is true, as this just says:

$\map {D_x} {\map {f_1} x} = \map {D_x} {\map {f_1} x}$

### Basis for the Induction

$\map P 2$ is the case:

$\map {D_x} {\map {f_1} x \map {f_2} x} = \map {D_x} {\map {f_1} x} \map {f_2} x + \map {f_1} x \map {D_x} {\map {f_2} x}$

which has been proved in Product Rule for Derivatives.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\ds \map {D_x} {\prod_{i \mathop = 1}^k \map {f_i} x} = \sum_{i \mathop = 1}^k \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$

Then we need to show:

$\ds \map {D_x} {\prod_{i \mathop = 1}^{k + 1} \map {f_i} x} = \sum_{i \mathop = 1}^{k + 1} \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$

### Induction Step

This is our induction step:

 $\ds \map {D_x} {\prod_{i \mathop = 1}^{k + 1} \map {f_i} x}$ $=$ $\ds \map {D_x} {\paren {\prod_{i \mathop = 1}^k \map {f_i} x} \map {f_{k + 1} } x}$ $\ds$ $=$ $\ds \map {D_x} {\map {f_{k + 1} } x} \paren {\prod_{i \mathop = 1}^k \map {f_i} x} + \map {D_x} {\prod_{i \mathop = 1}^k \map {f_i} x} \map {f_{k + 1} } x$ Basis for the Induction $\ds$ $=$ $\ds \map {D_x} {\map {f_{k + 1} } x} \paren {\prod_{i \mathop = 1}^k \map {f_i} x} + \paren {\sum_{i \mathop = 1}^k \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x} } \map {f_{k + 1} } x$ Induction Hypothesis $\ds$ $=$ $\ds \sum_{i \mathop = 1}^{k + 1} \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$ for all $n \in \N$

$\blacksquare$

## Mnemonic device

$\paren {f g}' = f g' + f' g$
$\paren {f g h}' = f' g h + f g' h + f g h'$

and in general, making sure to exhaust all possible combinations, making sure that there are as many summands as there are functions being multiplied.