Derivative of a Sum (x^n, ax^n)

We just looked at our first of many rules for differentiation (the process of finding the derivative).

Adding together terms in varying power forms create other functions, which in turn we can find the derivative of.  

For example if we have the functions $h(x)=2x$h(x)=2x and $g(x)=x^3$g(x)=x3 then we can add $h(x)$h(x) and $g(x)$g(x) together to create a new function $f(x)=x^3+2x$f(x)=x3+2x. 

To see what happens with the sum of the function and the resulting gradient function, let's look at 3 parts. Firstly $h(x)$h(x), then $g(x)$g(x) and then $f(x)$f(x).

$h(x)=2x$h(x)=2x then $h'(x)=2$h′(x)=2

for $g(x)=x^3$g(x)=x3 then $g'(x)=3x^2$g′(x)=3x2

and finally for $f(x)=x^3+2x$f(x)=x3+2x

Well, we are not sure yet how to find this so for the last one we will use first principles.

So we need to know $f(x)$f(x) and $f(x+h)$f(x+h)

$f(x)=x^3+2x$f(x)=x3+2x

$f(x+h)=(x+h)^3+2(x+h)$f(x+h)=(x+h)3+2(x+h)

$f(x+h)=h^3+3h^2x+3hx^2+x^3+2x+2h$f(x+h)=h3+3h2x+3hx2+x3+2x+2h

$f(x+h)=x^3+2x+h^3+3h^2x+3hx^2+2h$f(x+h)=x3+2x+h3+3h2x+3hx2+2h

$f(x+h)=[h^3+3h^2x+3hx^2+2h]+[x^3+2x]$f(x+h)=[h3+3h2x+3hx2+2h]+[x3+2x]
 

$\frac{dy}{dx}$dydx​	$=$=
$\frac{dy}{dx}$dydx​	$=$=
$\frac{dy}{dx}$dydx​	$=$=
$\frac{dy}{dx}$dydx​	$=$=
$\frac{dy}{dx}$dydx​	$=$=
$\frac{dy}{dx}$dydx​	$=$=	$3x^2+2$3x2+2

 

And this applies to any function, whether it be the $x^n$xn we saw before, or $ax^n$axn that we are exploring now.  

Examples

Find the derivative of the following, 

a) $f(x)=4x^2+3x+2$f(x)=4x2+3x+2, then $f'(x)=8x+3$f′(x)=8x+3   (remember that the derivative of a constant term is $0$0)

b) $f(x)=3x^3-3x^2$f(x)=3x3−3x2,  then $f'(x)=9x^2-6x$f′(x)=9x2−6x

c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x−3−2x+√x.  Firstly we need to turn the $\sqrt{x}$√x into a power.  $\sqrt{x}=x^{\frac{1}{2}}$√x=x12​ So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x−3−2x+x12​ and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f′(x)=−18x−4−2+12​x−12​

Worked Examples:

question 1

question 2