# Sum and Difference Rules

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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))'=lim_(h->0)((f(x+h)+g(x+h))-(f(x)+g(x)))/h=

=lim_(h->0)((f(x+h)-f(x))/h+(g(x+h)-g(x))/h)=

=lim_(h->0)(f(x+h)-f(x))/h+lim_(h->0)(g(x+h)-g(x))/h=

=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)=2x^6-3x^5+7x^3-2x^2+5

f'(x)=(2x^6-3x^5+7x^3-2x^2+5)'=(2x^6)'-(3x^5)'+(7x^3)'-(2x^2)'+(5)'=

=2(x^6)'-3(x^5)'+7(x^3)'-2(x^2)'+0=2*6x^(6-1)-3*5x^(5-1)+7*3x^(3-1)-2*2x^(2-1)=

=12x^5-15x^4+21x^2-4x.

Example 2. Find the points where f(x)=x^3-12x+23 has horizontal tangent lines.

Horizontal tangent lines are lines with slope 0, this will happen when derivative is zero.

f'(x)=(x^3-12x+23)'=(x^3)'-(12x)'+(23)'=(x^3)'-12(x)'+0=

=3x^(3-1)-12*1x^(1-1)=3x^2-12 .

f'(x)=0 when 3x^2-12=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)=3s^4-5s^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'(t)=(3s^4-5s^3+s)'=(3s^4)-(5s^3)'+(s)'=3(s^4)'-5(s^3)'+(s)'=

=3*4s^(4-1)-5*3s^(3-1)+1*s^(1-1)=12s^3-15s^2+1.

Thus, s'(1)=12*1^3-15*1^2+1=-2. Therefore velocity after 1 seconds is -2 m/s.