Let $f$ and $g$ be functions defined on an interval $A$, and assume both are partialerentiable at some point $c \in A$. Then,
$\left. (f + g) \right.'\left. (c) \right. = f'\left. (c) \right. + g'\left. (c) \right.$
$\left. (kf) \right.'\left. (c) \right. = kf'\left. (c) \right.$ for all $k \in \mathbb{R}$
$\left. (fg) \right.'\left. (c) \right. = f'\left. (c) \right.g\left. (c) \right. + f\left. (c) \right.g'\left. (c) \right.$
$\left. (f/g) \right.'\left. (c) \right. = \frac{g\left. (c) \right.f'\left. (c) \right. - f\left. (c) \right.g'\left. (c) \right.}{\left. \left\lbrack g\left. (c) \right. \right\rbrack \right.^{2}}$, provided that $g\left. (c) \right. \neq 0$
Proof of (1)#
$$
\begin{aligned}
\left. (f + g) \right.'\left. (c) \right. & = \lim\limits_{x \rightarrow c}\frac{\left. (f + g) \right.\left. (x) \right. - \left. (f + g) \right.\left. (c) \right.}{x - c} \\
& = \lim\limits_{x \rightarrow c}\frac{f\left. (x) \right. + g\left. (x) \right. - f\left. (c) \right. + g\left. (c) \right.}{x - c} \\
& = \lim\limits_{x \rightarrow c}\frac{\left. \left\lbrack f\left. (x) \right. - f\left. (c) \right. \right\rbrack \right. + \left. \left\lbrack g\left. (x) \right. - g\left. (c) \right. \right\rbrack \right.}{x - c} \\
& = \lim\limits_{x \rightarrow c}\frac{f\left. (x) \right. - f\left. (c) \right.}{x - c} + \lim\limits_{x \rightarrow c}\frac{g\left. (x) \right. - g\left. (c) \right.}{x - c} \\
& = f'\left. (c) \right. + g'\left. (c) \right.
\end{aligned}
$$
Proof of (2)#
$$
\begin{aligned}
\left. (kf) \right.'\left. (c) \right. & = \lim\limits_{x \rightarrow c}\frac{\left. (kf) \right.\left. (x) \right. - \left. (kf) \right.\left. (c) \right.}{x - c} \\
& = \lim\limits_{x \rightarrow c}\frac{kf\left. (x) \right. - kf\left. (c) \right.}{x - c} \\
& = k\lim\limits_{x \rightarrow c}\frac{f\left. (x) \right. - f\left. (c) \right.}{x - c} \\
& = kf'\left. (c) \right.
\end{aligned}
$$