Theorem: If $f$ and $g$ are differentiable then $\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$
Proof: Again this is just a matter of applying the definition of $\frac{d}{dx}$ and then applying the Sum Law for limits.

Let $F(x)=f(x)+g(x)$. Then

$\frac{d}{dx}(F(x))=\lim_{h\to 0}\ \frac{1}{h}(F(x+h)\ -\ F(x))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)+g(x+h)\ -\ (f(x)+g(x)))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)\ -\ f(x)+g(x+h)\ -\ g(x))$

$=\lim_{h\to 0}(\frac{1}{h}(f(x+h)\ -\ f(x))+\frac{1}{h}(g(x+h)\ -\ g(x)))$

$=\lim_{h\to 0}\ \frac{1}{h}(f(x+h)\ -\ f(x))+\lim_{h\to 0}\ \frac{1}{h}(g(x+h)\ -\ g(x))$

$=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$