Power rule (x^n and ax^n)

Functions of the type $x^n$xn and $ax^n$axnare called power functions, and the power rule is a shortcut way to find the derivative without having to use first principles every time. 

But, we will use first principles to determine the rule.  (we will show the proof for $x^n$xn here, but if you want to see the more complicated one for $ax^n$axn, you can see it here)  

$f(x)$f(x)	$=$=	$x^n$xn
$f(x+h)$f(x+h)	$=$=	$(x+h)^n$(x+h)n
$f'(x)$f′(x)	$=$=	$\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$limh→0(f(x+h)−f(x)h​)
$f'(x)$f′(x)	$=$=	$\lim_{h\to0}\left(\frac{\left(x+h\right)^n-x^n}{h}\right)$limh→0((x+h)n−xnh​)

 

From here we need to work through some pretty intense algebra, but it distills down to an understanding of the expansion of $x^n-a^n$xn−an

$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+a^3x^{n-4}+......+a^{n-1})$xn−an=(x−a)(xn−1+axn−2+a2xn−3+a3xn−4+......+an−1)

Using this expansion we can also determine the expansion for $(x+h)^n-x^n$(x+h)n−xn

$(x+h)^n-x^n==[(x+h)-x][(x+h)^{n-1}+x(x+h)^{n-2}+x^2(x+h)^{n-3}+x^3(x+h)^{n-4}+......+x^{n-1}]$(x+h)n−xn==[(x+h)−x][(x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+......+xn−1]

$f'(x)$f′(x)	$=$=	$\lim_{h\to0}\left(\frac{h\left[\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right]}{h}\right)$limh→0(h[(x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+...xn−1]h​)
$f'(x)$f′(x)	$=$=	$\lim_{h\to0}\left(\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right)$limh→0((x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+...xn−1)
$f'(x)$f′(x)	$=$=	$x^{n-1}+x\left(x\right)^{n-2}+x^2\left(x\right)^{n-3}+x^3\left(x\right)^{n-4}+...x^{n-1}$xn−1+x(x)n−2+x2(x)n−3+x3(x)n−4+...xn−1
$f'(x)$f′(x)	$=$=	$x^{n-1}+\frac{x\left(x\right)^n}{x^2}+\frac{x^2\left(x\right)^n}{x^3}+\frac{x^3\left(x\right)^n}{x^4}+...x^{n-1}$xn−1+x(x)nx2​+x2(x)nx3​+x3(x)nx4​+...xn−1
$f'(x)$f′(x)	$=$=	$x^{n-1}+\frac{x^n}{^x}+\frac{x^n}{^x}+\frac{x^n}{^x}+...x^{n-1}$xn−1+xnx​+xnx​+xnx​+...xn−1
$f'(x)$f′(x)	$=$=	$x^{n-1}+x^{n-1}+x^{n-1}+x^{n-1}+...x^{n-1}$xn−1+xn−1+xn−1+xn−1+...xn−1
$f'(x)$f′(x)	$=$=	$nx^{n-1}$nxn−1

Now of course, that was a lot of algebraic gymnastics, check with your teacher you may or may not need to follow that entire proof.  For those that don't here is the key message. 

That for functions of the form $f(x)=x^n$f(x)=xn the derivative is $f'(x)=nx^{n-1}$f′(x)=nxn−1 and this is called the power rule. 

Examples

Find the derivatives of

a)$f(x)=x^2$f(x)=x2, $f'(x)=2x$f′(x)=2x

b) $g(m)=m^4$g(m)=m4, so $g'(m)=4m^3$g′(m)=4m3

 

 

There isn't a lot of difference between the functions $x^2$x2 and $3x^2$3x2 for example, except that we know the $3$3 has the effect of dilating the graph (in this case making it steeper).  This means that the value of the $3$3 must have an affect on the derivative of the graph. 

Functions of the type $ax^n$axn are also called power functions, a slight variation of the power rule can be used to find the derivative. 

For functions of the form $f(x)=ax^n$f(x)=axn the derivative is $f'(x)=nax^{n-1}$f′(x)=naxn−1 and this is also called the power rule. 

Examples

Find the derivatives of

a) $h(t)=-3t^{\frac{3}{2}}$h(t)=−3t32​, so $h'(t)=\frac{-9}{2}t^{\frac{1}{2}}$h′(t)=−92​t12​

Be careful with these fractional ones - they are a common source of student errors. 

b) $g(x)=\frac{2}{x^4}$g(x)=2x4​, firstly we need the function in power form, so we convert it and get that $g(x)=2x^{-4}$g(x)=2x−4.  Now we can use the power rule and see that $g'(x)=-8x^{-5}$g′(x)=−8x−5   Remember that $-4-1=-5$−4−1=−5.  A common mistake here is to arrive at a power of $-3$−3.  

 

 

 

 

Examples

Question 1

Question 2

Question 3