Derivative Sum Difference Proof

Original Function

$f(x)=p(x)+q(x)$

Proof

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ $f^{\prime }(x)={\lim }_{h\to 0}\frac{p(x+h)-p(x)+q(x+h)-q(x)}{h}$ $f^{\prime }(x)={\lim }_{h\to 0}\frac{p(x+h)-p(x)}{h}+\frac{q(x+h)-q(x)}{h}$ $f^{\prime }(x)={\lim }_{h\to 0}\frac{p(x+h)-p(x)}{h}+{\lim }_{h\to 0}\frac{q(x+h)-q(x)}{h}$ $f^{\prime }(x)=p^{\prime }(x)+q^{\prime }(x)$