# Sum and Difference Rules

## Related calculator: Online Derivative Calculator with Steps

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 $-{2}\frac{{m}}{{s}}$.