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

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

$$P(1)$$ 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:


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

Then we need to show:


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

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