The Sum Rule. If f and g are both differentiable then (f(x)+g(x))′=(f(x))′+(g(x))′.
Proof. By definition (f(x)+g(x))′=limh→0h(f(x+h)+g(x+h))−(f(x)+g(x))=
=limh→0(hf(x+h)−f(x)+hg(x+h)−g(x))=
=limh→0hf(x+h)−f(x)+limh→0hg(x+h)−g(x)=
=f′(x)+g′(x).
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′.
By writing f−g as f+(−1)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 (f(x)−g(x))′=(f(x))′−(g(x))′
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)=2x6−3x5+7x3−2x2+5
f′(x)=(2x6−3x5+7x3−2x2+5)′=(2x6)′−(3x5)′+(7x3)′−(2x2)′+(5)′=
=2(x6)′−3(x5)′+7(x3)′−2(x2)′+0=2⋅6x6−1−3⋅5x5−1+7⋅3x3−1−2⋅2x2−1=
=12x5−15x4+21x2−4x.
Example 2. Find the points where f(x)=x3−12x+23 has horizontal tangent lines.
Horizontal tangent lines are lines with slope 0, this will happen when derivative is zero.
f′(x)=(x3−12x+23)′=(x3)′−(12x)′+(23)′=(x3)′−12(x)′+0=
=3x3−1−12⋅1x1−1=3x2−12.
f′(x)=0 when 3x2−12=0 or x=±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(t)=3s4−5s3+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′(t)=(3s4−5s3+s)′=(3s4)−(5s3)′+(s)′=3(s4)′−5(s3)′+(s)′=
=3⋅4s4−1−5⋅3s3−1+1⋅s1−1=12s3−15s2+1.
Thus, s′(1)=12⋅13−15⋅12+1=−2. Therefore velocity after 1 seconds is −2sm.