Quotient Rule for Derivatives

Theorem
Let $j \left({x}\right), k \left({x}\right)$ 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.

Define the real function $f$ on $I$ by:


 * $\displaystyle f \left({x}\right) = \begin{cases}

\dfrac {j \left({x}\right)} {k \left({x}\right)} & : k \left({x}\right) \ne 0 \\ 0 & : \text{otherwise} \end{cases}$

Then, if $k \left({\xi}\right) \ne 0$, $f$ is differentiable at $\xi$, and furthermore:


 * $f^{\prime} \left({\xi}\right) = \dfrac {j^{\prime} \left({\xi}\right) k \left({\xi}\right) - j \left({\xi}\right) k^{\prime} \left({\xi}\right)} {k \left({\xi}\right)^2}$

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


 * $\displaystyle \forall x \in I: k \left({x}\right) \ne 0 \implies f^{\prime} \left({x}\right) = \frac {j^{\prime} \left({x}\right) k \left({x}\right) - j \left({x}\right) k^{\prime} \left({x}\right)} {\left({k \left({x}\right)}\right)^2}$

Proof
Let $\xi$ be such that $k \left({\xi}\right) \ne 0$.

From Differentiable Function is Continuous‎, $k$ is continuous at $\xi$.

It follows that there exists an $\epsilon > 0$, such that $\left|{h}\right| < \epsilon \implies k \left({\xi + h}\right) \ne 0$.

So let $\left|{h}\right| < \epsilon$.

Then we have:

Hence:

Thus by:
 * continuity of $k$ at $\xi$
 * differentiability of $j$ and $k$ at $\xi$
 * Combined Sum Rule for Limits of Functions:

it is concluded that:
 * $\displaystyle \lim_{h \mathop \to 0} \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h = \frac 1 {k \left({\xi}\right)^2} \left({j^\prime \left({\xi}\right) k \left({\xi}\right) - j \left({\xi}\right) k^\prime \left({\xi}\right)}\right)$

From the definition of differentiability, $f$ is differentiable at $\xi$, with stated value.