Quotient Rule for Derivatives

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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.

$\blacksquare$


Examples

Example: $\dfrac x {x + 1}$

$\map {\dfrac \d {\d x} } {\dfrac x {x + 1} } = \dfrac 1 {\paren {x + 1}^2}$


Example: $\dfrac {x + 1} {x - 1}$

$\map {\dfrac \d {\d x} } {\dfrac {x + 1} {x - 1} } = -\dfrac 2 {\paren {x - 1}^2}$


Example: $\dfrac x {\cos x}$

$\map {\dfrac \d {\d x} } {\dfrac x {\cos x} } = \dfrac {\cos x + x \sin x} {\cos^2 x}$


Example: $\dfrac {e^x} x$

$\map {\dfrac \d {\d x} } {\dfrac {e^x} x} = \dfrac {e^x \paren {x - 1} } {x^2}$


Example: $\dfrac {\paren {x - 1} \paren {2 x - 1} } {x - 2}$

$\map {\dfrac \d {\d x} } {\dfrac {\paren {x - 1} \paren {2 x - 1} } {x - 2} } = \dfrac {2 x^2 - 8 x + 5} {\paren {x - 2}^2}$


Sources

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