Sum 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 $f$ is differentiable at $\xi$ and:

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


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

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


General Result

Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions all differentiable.

Then for all $n \in \N_{>0}$:

$\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$


Proof 1

\(\ds \map {f'} \xi\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h\) Definition of Derivative
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} + \map k {\xi + h} } - \paren {\map j \xi + \map k \xi } } h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} + \map k {\xi + h} - \map j \xi - \map k \xi} h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} - \map j \xi} + \paren {\map k {\xi + h} - \map k \xi} } h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h + \frac {\map k {\xi + h} - \map k \xi} h}\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map j {\xi + h} - \map j \xi} h + \lim_{h \mathop \to 0} \frac {\map k {\xi + h} - \map k \xi} h\)
\(\ds \) \(=\) \(\ds \map {j'} \xi + \map {k'} \xi\) Definition of Derivative

$\blacksquare$


Proof 2

It can be observed that this is an example of a Linear Combination of Derivatives with $\lambda = \mu = 1$.

$\blacksquare$


Historical Note

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


Sources

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