Product Rule for Derivatives

Theorem
Let $f \left({x}\right), j \left({x}\right), k \left({x}\right)$ be real functions defined on the open interval $I$.

Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.

Let $f \left({x}\right) = j \left({x}\right) k \left({x}\right)$.

Then:
 * $f^{\prime} \left({\xi}\right) = j \left({\xi}\right) k^{\prime} \left({\xi}\right) + j^{\prime} \left({\xi}\right) k \left({\xi}\right)$.

It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:


 * $\forall x \in I: f^{\prime} \left({x}\right) = j \left({x}\right) k^{\prime} \left({x}\right) + j^{\prime} \left({x}\right) k \left({x}\right)$

General Result
Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be real functions all differentiable as above.

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

Proof
Note that $j \left({\xi + h}\right) \to j \left({\xi}\right)$ as $h \to 0$ because, from Differentiable Function is Continuous‎, $j$ is continuous at $\xi$.

Proof of General Result
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle D_x \left({\prod_{i=1}^n f_i \left({x}\right)}\right) = \sum_{i=1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_i \left({x}\right)}\right)$

$P(1)$ 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 above.

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:


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

Then we need to show:


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

Induction Step
This is our induction step:

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

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