### Theorem If $u$ and $v$ are differentiable functions of $x$, then their sum $u+v$ is differentiable at every point where $u$ and $v$ are both differentiable. At such points, $ \frac{d}{d x}(u+v)=\frac{d u}{d x}+\frac{d v}{d x} . $ #### Proof We apply the definition of the [[derivative]] to $f(x)=u(x)+v(x)$ : $ \begin{aligned} \frac{d}{d x}[u(x)+v(x)] &=\lim _{h \rightarrow 0} \frac{[u(x+h)+v(x+h)]-[u(x)+v(x)]}{h} \\ &=\lim _{h \rightarrow 0}\left[\frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h}\right] \\ &=\lim _{h \rightarrow 0} \frac{u(x+h)-u(x)}{h}+\lim _{h \rightarrow 0} \frac{v(x+h)-v(x)}{h}=\frac{d u}{d x}+\frac{d v}{d x} . \end{aligned} $