Sum and Difference Rules
The Sum Rule. If $$$f$$$ and $$$g$$$ are both differentiable then $$${\left({f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}\right)}'={\left({f{{\left({x}\right)}}}\right)}'+{\left({g{{\left({x}\right)}}}\right)}'$$$.
Proof. By definition $$${\left({f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}\right)}'=\lim_{{{h}\to{0}}}\frac{{{\left({f{{\left({x}+{h}\right)}}}+{g{{\left({x}+{h}\right)}}}\right)}-{\left({f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}\right)}}}{{h}}=$$$
$$$=\lim_{{{h}\to{0}}}{\left(\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}+\frac{{{g{{\left({x}+{h}\right)}}}-{g{{\left({x}\right)}}}}}{{h}}\right)}=$$$
$$$=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}+\lim_{{{h}\to{0}}}\frac{{{g{{\left({x}+{h}\right)}}}-{g{{\left({x}\right)}}}}}{{h}}=$$$
$$$={f{'}}{\left({x}\right)}+{g{'}}{\left({x}\right)}$$$.
In general, sum rule can be extended to the sum of any number of functions. Indeed, $$${\left({f{+}}{g{+}}{h}\right)}'={\left({f{+}}{\left({g{+}}{h}\right)}\right)}'={f{'}}+{\left({g{+}}{h}\right)}'={f{'}}+{g{'}}+{h}'$$$.
By writing $$${f{-}}{g{}}$$$ as $$${f{+}}{\left(-{1}\right)}{g{}}$$$ and applying sum rule and constant multiple rule, we get the following formula:
The Difference Rule. If $$${f{}}$$$ and $$${g{}}$$$ are both differentiable then $$${\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}'={\left({f{{\left({x}\right)}}}\right)}'-{\left({g{{\left({x}\right)}}}\right)}'$$$
Using combinations of derivative of power function, constant multiple, sum and difference rules, we can find derivative of any polynomial.
Example 1. Differentiate $$${f{{\left({x}\right)}}}={2}{{x}}^{{6}}-{3}{{x}}^{{5}}+{7}{{x}}^{{3}}-{2}{{x}}^{{2}}+{5}$$$
$$${f{'}}{\left({x}\right)}={\left({2}{{x}}^{{6}}-{3}{{x}}^{{5}}+{7}{{x}}^{{3}}-{2}{{x}}^{{2}}+{5}\right)}'={\left({2}{{x}}^{{6}}\right)}'-{\left({3}{{x}}^{{5}}\right)}'+{\left({7}{{x}}^{{3}}\right)}'-{\left({2}{{x}}^{{2}}\right)}'+{\left({5}\right)}'=$$$
$$$={2}{\left({{x}}^{{6}}\right)}'-{3}{\left({{x}}^{{5}}\right)}'+{7}{\left({{x}}^{{3}}\right)}'-{2}{\left({{x}}^{{2}}\right)}'+{0}={2}\cdot{6}{{x}}^{{{6}-{1}}}-{3}\cdot{5}{{x}}^{{{5}-{1}}}+{7}\cdot{3}{{x}}^{{{3}-{1}}}-{2}\cdot{2}{{x}}^{{{2}-{1}}}=$$$
$$$={12}{{x}}^{{5}}-{15}{{x}}^{{4}}+{21}{{x}}^{{2}}-{4}{x}$$$.
Example 2. Find the points where $$${f{{\left({x}\right)}}}={{x}}^{{3}}-{12}{x}+{23}$$$ has horizontal tangent lines.
Horizontal tangent lines are lines with slope 0, this will happen when derivative is zero.
$$${f{'}}{\left({x}\right)}={\left({{x}}^{{3}}-{12}{x}+{23}\right)}'={\left({{x}}^{{3}}\right)}'-{\left({12}{x}\right)}'+{\left({23}\right)}'={\left({{x}}^{{3}}\right)}'-{12}{\left({x}\right)}'+{0}=$$$
$$$={3}{{x}}^{{{3}-{1}}}-{12}\cdot{1}{{x}}^{{{1}-{1}}}={3}{{x}}^{{2}}-{12}$$$.
$$${f{'}}{\left({x}\right)}={0}$$$ when $$${3}{{x}}^{{2}}-{12}={0}$$$ or $$${x}=\pm{2}$$$.
So, points where tangent lines are horizontal are $$${x}={2}$$$ and $$${x}=-{2}$$$.
Example 3. The equation of motion of the particle is $$${s}{\left({t}\right)}={3}{{s}}^{{4}}-{5}{{s}}^{{3}}+{s}$$$ where $$${s}$$$ is measured in meters and $$${t}$$$ is seconds. Find velocity of particle after 1 second.
Velocity of particle is derivative of displacement.
$$${s}'{\left({t}\right)}={\left({3}{{s}}^{{4}}-{5}{{s}}^{{3}}+{s}\right)}'={\left({3}{{s}}^{{4}}\right)}-{\left({5}{{s}}^{{3}}\right)}'+{\left({s}\right)}'={3}{\left({{s}}^{{4}}\right)}'-{5}{\left({{s}}^{{3}}\right)}'+{\left({s}\right)}'=$$$
$$$={3}\cdot{4}{{s}}^{{{4}-{1}}}-{5}\cdot{3}{{s}}^{{{3}-{1}}}+{1}\cdot{{s}}^{{{1}-{1}}}={12}{{s}}^{{3}}-{15}{{s}}^{{2}}+{1}$$$.
Thus, $$${s}'{\left({1}\right)}={12}\cdot{{1}}^{{3}}-{15}\cdot{{1}}^{{2}}+{1}=-{2}$$$. Therefore velocity after 1 seconds is $$$-{2}\frac{{m}}{{s}}$$$.