# Sum Rule for Derivatives/Proof 1

## 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:

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

## Proof

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

$\blacksquare$