Quotient Rule for Derivatives

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

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

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

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 $$\xi$$.