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:
 * $f' \left({\xi}\right) = j \left({\xi}\right) \, k' \left({\xi}\right) + j\,' \left({\xi}\right) \, k \left({\xi}\right)$.

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


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

Using Leibniz's Notation for Derivatives, this can be written as:


 * $\dfrac{\mathrm d}{\mathrm dx}\left({y \, z}\right) = y \dfrac{\mathrm dz}{\mathrm dx} + \dfrac{\mathrm dy}{\mathrm dx} z$

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

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 $\xi$.

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