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:

$\displaystyle f \left({x}\right) = \begin{cases} \dfrac {j \left({x}\right)} {k \left({x}\right)} & \text{if } k \left({x}\right) \ne 0 \\ 0 & \text{otherwise} \end{cases}$

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

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

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 suppose $\left|{h}\right| < \epsilon$. Then we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h\)	\(=\)	\(\displaystyle \frac 1 h \left({\frac{j \left({\xi + h}\right)}{k \left({\xi + h}\right)} - \frac {j \left({\xi}\right)}{k \left({\xi}\right)} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Since $k \left({\xi}\right), k \left({\xi + h}\right) \ne 0$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {j \left({\xi + h}\right) k \left({\xi}\right) - k \left({\xi + h}\right) j \left({\xi}\right)} {h k \left({\xi + h}\right) k \left({\xi}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \left({ \frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{ k \left({\xi + h}\right) - k \left({\xi}\right) } h }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Algebraic manipulations
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \lim_{h \to 0} \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h\)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \left({ \frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{ k \left({\xi + h}\right) - k \left({\xi}\right) } h }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac 1 {k \left({\xi + h}\right) k \left({\xi}\right)} \cdot \lim_{h \to 0} \left({ \frac{j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right) - j \left({\xi}\right) \frac{ k \left({\xi + h}\right) - k \left({\xi}\right) } h }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Product Rule for Limits of Functions
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {k \left({\xi}\right)^2} \left({j^\prime \left({\xi}\right) k \left({\xi}\right) - j \left({\xi}\right) k^\prime \left({\xi}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Continuity of $k$ at $\xi$; differentiability of $j$ and $k$ at $\xi$; and Combined Sum Rule for Limits of Functions

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

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.9$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 10.9 \ \text{(iii)}$