# Product Rule for Derivatives/General Result

## Theorem

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

then:

$\forall x \in I: \displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$

## Proof

Proof by induction:

For all $n \in \N_{\ge 1}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$

$P(1)$ is true, as this just says:

$D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$

### Basis for the Induction

$P(2)$ is the case:

$D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right)$

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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$

Then we need to show:

$\displaystyle D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$

### Induction Step

This is our induction step:

 $\displaystyle D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right)$ $=$ $\displaystyle D_x \left({\left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + \left({\sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$

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

Therefore:

$\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ for all $n \in \N$

$\blacksquare$

## Mnemonic device

$\left({f g}\right)' = f g' + f' g$
$\left({f g h}\right)' = f' g h + f g' h + f g h'$

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