Product Rule for Derivatives/Proof

Theorem

Let $\map f x, \map j x, \map k x$ 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 $\map f x = \map j x \map k x$.

Then:

$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$

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

$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$

Proof

First we note that from Differentiable Function is Continuousâ€Ž, $j$ is continuous at $\xi$.

Hence:

$(1): \quad \map j {\xi + h} \to \map j \xi$ as $h \to 0$

So:

 $\ds \map {f'} \xi$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h$ Definition of Derivative $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j \xi \map k \xi} h$ by hypothesis $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j {\xi + h} \map k \xi + \map j {\xi + h} \map k \xi - \map j \xi \map k \xi} h$ adding $\pm \map j {\xi + h} \map k \xi$ to numerator $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h + \frac {\map j {\xi + h} - \map j \xi} h \map k \xi}$ simplifying $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h} + \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi}$ Sum Rule for Limits of Real Functions $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} } \lim_{h \mathop \to 0} \paren {\frac {\map k {\xi + h} - \map k \xi} h} + \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h} \lim_{h \mathop \to 0} \paren {\map k \xi}$ Product Rule for Limits of Real Functions $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} } \map {k'} \xi + \map {j'} \xi \lim_{h \mathop \to 0} \paren {\map k \xi}$ Definition of Derivative $\ds$ $=$ $\ds \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$ from $(1)$

$\blacksquare$