9.04 Differentiation of y=x^n

Functions of the type $f(x)=x^n$f(x)=xn are called power functions. From our last lesson on first principles you may have begun to notice some patterns in the derivatives found, this leads to developing rules to differentiate certain functions rather than using first principles each time. Remember that the derivative gives the gradient of the tangent to the curve. 

• The derivative of a constant is zero. (E.g. the derivative of $f(x)=5$f(x)=5 is $f'(x)=0$f′(x)=0).
  We actually knew this already because we know that the graph of a constant function is a horizontal line, which has gradient of $0$0.

Practice question

Question 1

 

• The derivative of a linear function is a constant. (E.g. the derivative of $f(x)=3x+1$f(x)=3x+1 is $f'(x)=3$f′(x)=3).
  We actually knew this already because we know that a function in the form $y=mx+c$y=mx+c has gradient of $m$m. 

Practice question

Question 2

 

What we didn't already know:

• The derivative of a quadratic function is a linear function. (E.g. the derivative of $f(x)=4x^2+3x+1$f(x)=4x2+3x+1 is $f'(x)=8x+3$f′(x)=8x+3)
• The derivative of a cubic function is a quadratic function. (E.g. the derivative of $f(x)=x^3+6x$f(x)=x3+6x is $f'(x)=3x^2+6$f′(x)=3x2+6)
• The derivative of a quartic function is a cubic function. (E.g. the derivative of $f(x)=2x^4$f(x)=2x4 is $f'(x)=8x^3$f′(x)=8x3)

This suggests that the derivative of a polynomial of degree $n$n is of degree $n-1$n−1, for integer values of $n\ge1$n≥1.

Looking further at just examples of functions of the form $f(x)=x^n$f(x)=xn, we have:

$f(x)$f(x)	$f'(x)$f′(x)
$x$x	$1$1
$x^2$x2	$2x$2x
$x^3$x3	$3x^2$3x2
$x^4$x4	$4x^3$4x3
$x^5$x5	$5x^4$5x4

Noticing the pattern in the table, we could make the conjecture that the derivative of $f(x)=x^n$f(x)=xn is $f'(x)=nx^{n-1}$f′(x)=nxn−1. This is in fact the case and we call this formula the power rule. The power rule applies to not only positive integer values of $n$n but for $n$n being any real number. Once we establish the rule we can use the power rule as a shortcut to find the derivative without having to use first principles every time. 

Let's try applying this rule and then look in detail of how to establish this rule using first principles.

 

Worked examples

Example 1

Find the derivatives of the following functions using the power rule.

Think: The power rule tells us "bring the power to the front and then subtract one from the power".

(a) $f(x)=x^2$f(x)=x2

Thus, $f'(x)=2x$f′(x)=2x.

(b) $g(m)=m^4$g(m)=m4

Thus, $g'(m)=4m^3$g′(m)=4m3.

(c) $h(t)=t^{\frac{3}{2}}$h(t)=t32​

Thus,

$h'(t)$h′(t)	$=$=	$\frac{3}{2}t^{\frac{3}{2}-1}$32​t32​−1
	$=$=	$\frac{3}{2}t^{\frac{1}{2}}$32​t12​

(d) $g(x)=\frac{1}{x^4}$g(x)=1x4​

Think: Firstly we need the function in power form, recall from index laws that $\frac{1}{x^n}=x^{-n}$1xn​=x−n, so we have $g(x)=x^{-4}$g(x)=x−4.  

Do: Now we can use the power rule and see that $g'(x)=-4x^{-5}$g′(x)=−4x−5. Take care with negative powers, remember that $-4-1=-5$−4−1=−5.

Example 2

Find the derivative of $f(x)=x^2\sqrt{x}$f(x)=x2√x and hence, find the gradient of the tangent to the function at $x=4$x=4.

Think: First use index laws to write the function in the form $f(x)=x^n$f(x)=xn. 

$f(x)$f(x)	$=$=	$x^2\sqrt{x}$x2√x
	$=$=	$x^2\times x^{\frac{1}{2}}$x2×x12​
	$=$=	$x^{\frac{5}{2}}$x52​

Do: Now we can use the power rule to find the derivative.

$f'(x)$f′(x)	$=$=	$\frac{5}{2}x^{\frac{5}{2}-1}$52​x52​−1
	$=$=	$\frac{5}{2}x^{\frac{3}{2}}$52​x32​

Lastly, we were asked to find the gradient of the tangent at $x=4$x=4, so evaluate $f'(4)$f′(4):

$f'(4)$f′(4)	$=$=	$\frac{5}{2}\left(4\right)^{\frac{3}{2}}$52​(4)32​
	$=$=	$20$20

Proof of the power rule

We will look at proving the power rule for positive integers $n$n using first principles. To prove the more general case requires knowledge for differentiation of exponential functions which you will encounter in year $12$12. 

Using first principles for a function $f(x)=x^n$f(x)=xn, the derivative $f'(x)$f′(x) is defined to be:

$f'(x)$f′(x)	$=$=	$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$limh→0f(x+h)−f(x)h​
	$=$=	$\lim_{h\rightarrow0}\frac{\left(x+h\right)^n-x^n}{h}$limh→0(x+h)n−xnh​

From here we have two choices for how to move forward, we can expand $\left(x+h\right)^n-x^n$(x+h)n−xn or we can factorise it. Let's look at both briefly and we should arrive at the same conclusion.

Method 1 - expansion

To expand $\left(x+h\right)^n-x^n$(x+h)n−xn, recall the binomial theorem, we will get something of the form:

$\left(x+h\right)^n-x^n=x^n+nhx^{n-1}+...+nh^{n-1}x+h^n-x^n$(x+h)n−xn=xn+nhxn−1+...+nhn−1x+hn−xn

As the central terms of the expansion will all contain a factor of $h$h when the limit is taken they will not impact our final result and so let's use this shorthand version.

Hence,

$f'(x)$f′(x)	$=$=	$\lim_{h\rightarrow0}\frac{(x+h)^n-x^n}{h}$limh→0(x+h)n−xnh​
	$=$=	$\lim_{h\rightarrow0}\frac{x^n+nhx^{n-1}+...+nh^{n-1}x+h^n-x^n}{h}$limh→0xn+nhxn−1+...+nhn−1x+hn−xnh​
	$=$=	$\lim_{h\rightarrow0}\frac{nhx^{n-1}+...+nh^{n-1}x+h^n}{h}$limh→0nhxn−1+...+nhn−1x+hnh​	The first and last term of the expansion cancelled out
	$=$=	$\lim_{h\rightarrow0}\frac{h(nx^{n-1}+...nh^{n-2}x+h^{n-1})}{h}$limh→0h(nxn−1+...nhn−2x+hn−1)h​	Factor out an $h$h from each term in the numerator
	$=$=	$\lim_{h\rightarrow0}(nx^{n-1}+...+nh^{n-2}x+h^{n-1})$limh→0(nxn−1+...+nhn−2x+hn−1)	Divide by $h$h
	$=$=	$nx^{n-1}$nxn−1	Take the limit by letting $h=0$h=0 and all but the first term will disappear

 

Method 2 - factorisation

For this we require the knowledge that $a^n-b^n$an−bn factorises as:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+a^{n-4}b^3+......+ab^{n-2}+b^{n-1})$an−bn=(a−b)(an−1+an−2b+an−3b2+an−4b3+......+abn−2+bn−1)

Using this factorisation we can determine a new form for $(x+h)^n-x^n$(x+h)n−xn:

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

Hence,

$f'(x)$f′(x)	$=$=	$\lim_{h\rightarrow0}\frac{(x+h)^n-x^n}{h}$limh→0(x+h)n−xnh​
	$=$=	$\lim_{h\to0}\left(\frac{h\left[\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right]}{h}\right)$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​)
	$=$=	$\lim_{h\to0}\left(\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right)$limh→0((x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+...xn−1)	Divide by $h$h
	$=$=	$x^{n-1}+x\left(x\right)^{n-2}+x^2\left(x\right)^{n-3}+x^3\left(x\right)^{n-4}+...x^{n-1}$xn−1+x(x)n−2+x2(x)n−3+x3(x)n−4+...xn−1	Take the limit by letting $h=0$h=0
	$=$=	$x^{n-1}+x^{n-2+1}+x^{n-3+2}+x^{n-4+3}+...x^{n-1}$xn−1+xn−2+1+xn−3+2+xn−4+3+...xn−1	Rewriting powers using index laws
	$=$=	$x^{n-1}+x^{n-1}+x^{n-1}+x^{n-1}+...x^{n-1}$xn−1+xn−1+xn−1+xn−1+...xn−1	Simplify powers
	$=$=	$nx^{n-1}$nxn−1	Each power was the same and there were $n$nterms

 

Practice questions

Question 3

Question 4

Question 5

QUESTION 6