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:


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

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:

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 $n \in \N$

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.

Also see

 * Derivative of Product of Real Function and Vector-Valued Function
 * Derivative of Vector Cross Product of Vector-Valued Functions
 * Derivative of Dot Product of Vector-Valued Functions