Quotient Rule for Derivatives/Proof
Theorem
Let $\map j x, \map k x$ 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:
- $\ds \map f x = \begin{cases}
\dfrac {\map j x} {\map k x} & : \map k x \ne 0 \\ 0 & : \text{otherwise} \end{cases}$
Then, if $\map k \xi \ne 0$, $f$ is differentiable at $\xi$, and furthermore:
- $\map {f'} \xi = \dfrac {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi} {\paren {\map k \xi}^2}$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
- $\ds \forall x \in I: \map k x \ne 0 \implies \map {f'} x = \frac {\map {j'} x \map k x - \map j x \map {k'} x} {\paren {\map k x}^2}$
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:
\(\ds \frac {\map f {\xi + h} - \map f \xi} h\) | \(=\) | \(\ds \frac 1 h \paren {\frac {\map j {\xi + h} } {\map k {\xi + h} } - \frac {\map j \xi} {\map k \xi } }\) | as $\map k \xi, \map k {\xi + h} \ne 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map j {\xi + h} \map k \xi - \map k {\xi + h} \map j \xi} {h \map k {\xi + h} \map k \xi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}\) | algebraic manipulation |
Hence:
\(\ds \) | \(\) | \(\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \cdot \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}\) | Product Rule for Limits of Real Functions |
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.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Quotient
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 10.9 \ \text{(iii)}$