Quotient Rule for Derivatives/Proof

Proof
Let $\xi$ be such that $\map k \xi \ne 0$.

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

It follows that there exists an $\epsilon > 0$, such that $\size h < \epsilon \implies \map k {\xi + h} \ne 0$.

So let $\size h < \epsilon$.

Then we have:

Hence:

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

it is concluded that:
 * $\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h = \frac 1 {\map k \xi^2} \paren {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi}$

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