Quotient Rule for Derivatives

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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:

\(\displaystyle \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h\) \(=\) \(\displaystyle \frac 1 h \left({\frac{j \left({\xi + h}\right)}{k \left({\xi + h}\right)} - \frac {j \left({\xi}\right)}{k \left({\xi}\right)} }\right)\) Since $k \left({\xi}\right), k \left({\xi + h}\right) \ne 0$
\(\displaystyle \) \(=\) \(\displaystyle \frac {j \left({\xi + h}\right) k \left({\xi}\right) - k \left({\xi + h}\right) j \left({\xi}\right)} {h k \left({\xi + h}\right) k \left({\xi}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \left({ \frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{k \left({\xi + h}\right) - k \left({\xi}\right) } h }\right)\) algebraic manipulation


Hence:

\(\displaystyle \) \(\) \(\displaystyle \lim_{h \mathop \to 0} \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \left({ \frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{ k \left({\xi + h}\right) - k \left({\xi}\right) } h }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \cdot \lim_{h \to 0} \left({\frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{ k \left({\xi + h}\right) - k \left({\xi}\right) } h}\right)\) Product Rule for Limits of Functions


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.

$\blacksquare$


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