Sum and Difference Rules

The Sum Rule. If ff and gg are both differentiable then (f(x)+g(x))=(f(x))+(g(x)){\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 (f(x)+g(x))=limh0(f(x+h)+g(x+h))(f(x)+g(x))h={\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}}=

=limh0(f(x+h)f(x)h+g(x+h)g(x)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)}=

=limh0f(x+h)f(x)h+limh0g(x+h)g(x)h==\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(x)+g(x)={f{'}}{\left({x}\right)}+{g{'}}{\left({x}\right)}.

In general, sum rule can be extended to the sum of any number of functions. Indeed, (f+g+h)=(f+(g+h))=f+(g+h)=f+g+h{\left({f{+}}{g{+}}{h}\right)}'={\left({f{+}}{\left({g{+}}{h}\right)}\right)}'={f{'}}+{\left({g{+}}{h}\right)}'={f{'}}+{g{'}}+{h}'.

By writing fg{f{-}}{g{}} as f+(1)g{f{+}}{\left(-{1}\right)}{g{}} and applying sum rule and constant multiple rule, we get the following formula:

The Difference Rule. If f{f{}} and g{g{}} are both differentiable then (f(x)g(x))=(f(x))(g(x)){\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(x)=2x63x5+7x32x2+5{f{{\left({x}\right)}}}={2}{{x}}^{{6}}-{3}{{x}}^{{5}}+{7}{{x}}^{{3}}-{2}{{x}}^{{2}}+{5}

f(x)=(2x63x5+7x32x2+5)=(2x6)(3x5)+(7x3)(2x2)+(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(x6)3(x5)+7(x3)2(x2)+0=26x6135x51+73x3122x21=={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}}}=

=12x515x4+21x24x={12}{{x}}^{{5}}-{15}{{x}}^{{4}}+{21}{{x}}^{{2}}-{4}{x}.

Example 2. Find the points where f(x)=x312x+23{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(x)=(x312x+23)=(x3)(12x)+(23)=(x3)12(x)+0={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}=

=3x31121x11=3x212={3}{{x}}^{{{3}-{1}}}-{12}\cdot{1}{{x}}^{{{1}-{1}}}={3}{{x}}^{{2}}-{12}.

f(x)=0{f{'}}{\left({x}\right)}={0} when 3x212=0{3}{{x}}^{{2}}-{12}={0} or x=±2{x}=\pm{2}.

So, points where tangent lines are horizontal are x=2{x}={2} and x=2{x}=-{2}.

Example 3. The equation of motion of the particle is s(t)=3s45s3+s{s}{\left({t}\right)}={3}{{s}}^{{4}}-{5}{{s}}^{{3}}+{s} where s{s} is measured in meters and t{t} is seconds. Find velocity of particle after 1 second.

Velocity of particle is derivative of displacement.

s(t)=(3s45s3+s)=(3s4)(5s3)+(s)=3(s4)5(s3)+(s)={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)}'=

=34s4153s31+1s11=12s315s2+1={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(1)=12131512+1=2{s}'{\left({1}\right)}={12}\cdot{{1}}^{{3}}-{15}\cdot{{1}}^{{2}}+{1}=-{2}. Therefore velocity after 1 seconds is 2ms-{2}\frac{{m}}{{s}}.