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=`




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`




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.


`=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.



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