Product Rule for Derivatives

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Theorem

Let $\map f x, \map j x, \map k x$ 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.


Let $\map f x = \map j x \map k x$.


Then:

$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$


It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:

$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$


Using Leibniz's notation for derivatives, this can be written as:

$\map {\dfrac \d {\d x} } {y \, z} = y \dfrac {\d z} {\d x} + \dfrac {\d y} {\d x} z$

where $y$ and $z$ represent functions of $x$.


General Result

Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions differentiable on the open interval $I$.

then:

$\forall x \in I: \ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$


Proof

First we note that from Differentiable Function is Continuous‎, $j$ is continuous at $\xi$.

Hence:

$(1): \quad \map j {\xi + h} \to \map j \xi$ as $h \to 0$


So:

\(\ds \map {f'} \xi\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h\) Definition of Derivative
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j \xi \map k \xi} h\) by hypothesis
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j {\xi + h} \map k \xi + \map j {\xi + h} \map k \xi - \map j \xi \map k \xi} h\) adding $\pm \map j {\xi + h} \map k \xi$ to numerator
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h + \frac {\map j {\xi + h} - \map j \xi} h \map k \xi}\) simplifying
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h} + \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi}\) Sum Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} } \lim_{h \mathop \to 0} \paren {\frac {\map k {\xi + h} - \map k \xi} h} + \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h} \lim_{h \mathop \to 0} \paren {\map k \xi}\) Product Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \paren {\map j {\xi + h} } \map {k'} \xi + \map {j'} \xi \lim_{h \mathop \to 0} \paren {\map k \xi}\) Definition of Derivative
\(\ds \) \(=\) \(\ds \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi\) from $(1)$

$\blacksquare$


Examples

Example: $2 a x e^{a x^2}$

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


Example: $x \cot x$

$\map {\dfrac \d {\d x} } {x \cot x} = \cot x - x \cosec^2 x$


Example: $x^2 \arctan x$

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


Example: $x e^x \sin x$

$\map {\dfrac \d {\d x} } {x e^x \sin x} = e^x \paren {\paren {1 + x} \sin x + x \cos x}$


Also see


Historical Note

The Product Rule for Derivatives was first obtained by Gottfried Wilhelm von Leibniz in $1677$.


Sources

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