# Sum and Difference Rules

## Related Calculator: Online Derivative Calculator with Steps

**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`.