Quotient 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 $f \left({x}\right) = \dfrac {j \left({x}\right)} {k \left({x}\right)}$.

Then:
 * $\displaystyle f^{\prime} \left({\xi}\right) = \frac {j^{\prime} \left({\xi}\right) k \left({\xi}\right) - j \left({\xi}\right) k^{\prime} \left({\xi}\right)} {\left({k \left({\xi}\right)}\right)^2}$

provided $k \left({\xi}\right) \ne 0$.

It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
 * $\displaystyle \forall x \in I: f^{\prime} \left({x}\right) = \frac {j^{\prime} \left({x}\right) k \left({x}\right) - j \left({x}\right) k^{\prime} \left({x}\right)} {\left({k \left({x}\right)}\right)^2}$

wherever $k \left({x}\right) \ne 0$.

Proof
We have

Note that $k \left({\xi + h}\right) \to k \left({\xi}\right)$ as $h \to 0$ because, from Differentiable Function is Continuous‎, $k$ is continuous at $\xi$.