Product Rule for Derivatives

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:


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

Using Leibniz's notation for derivatives, this can be written as:


 * $\map {\dfrac \d {\d x} } {y \, z} = y \dfrac {\d z} {\d x} + \dfrac {\d y} {\d x} z$

where $y$ and $z$ represent functions of $x$.

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

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


 * Leibniz's Rule for One Variable, of which this is the special case of the first derivative