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8.04 Differentiation of y=x^n

Lesson

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. Did you notice:

  • 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)
  • 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)
  • 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 $n-1$n1, for integer values of $n\ge1$n1.

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)=nxn1. 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. 

Power rule

For a function $f(x)=x^n$f(x)=xn, the derivative is $f'(x)=nx^{n-1}$f(x)=nxn1, for $n$n any real number.

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}$32t321
  $=$= $\frac{3}{2}t^{\frac{1}{2}}$32t12

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=xn, so we have $g(x)=x^{-4}$g(x)=x4.  

Do: Now we can use the power rule and see that $g'(x)=-4x^{-5}$g(x)=4x5. Take care with negative powers, remember that $-4-1=-5$41=5.

Example 2

Find the derivative of $f(x)=x^2\sqrt{x}$f(x)=x2x 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) $=$= $x^2\sqrt{x}$x2x
  $=$= $x^2\times x^{\frac{1}{2}}$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}$52x521
  $=$= $\frac{5}{2}x^{\frac{3}{2}}$52x32

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 at 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) $=$= $\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$limh0f(x+h)f(x)h
  $=$= $\lim_{h\rightarrow0}\frac{\left(x+h\right)^n-x^n}{h}$limh0(x+h)nxnh

From here we have two choices for how to move forward, we can expand $\left(x+h\right)^n-x^n$(x+h)nxn 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)nxn, 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)nxn=xn+nhxn1+...+nhn1x+hnxn

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) $=$= $\lim_{h\rightarrow0}\frac{(x+h)^n-x^n}{h}$limh0(x+h)nxnh  
  $=$= $\lim_{h\rightarrow0}\frac{x^n+nhx^{n-1}+...+nh^{n-1}x+h^n-x^n}{h}$limh0xn+nhxn1+...+nhn1x+hnxnh  
  $=$= $\lim_{h\rightarrow0}\frac{nhx^{n-1}+...+nh^{n-1}x+h^n}{h}$limh0nhxn1+...+nhn1x+hnh The first and last term of the expansion cancelled out
  $=$= $\lim_{h\rightarrow0}\frac{h(nx^{n-1}+...nh^{n-2}x+h^{n-1})}{h}$limh0h(nxn1+...nhn2x+hn1)h Factor out a $h$h from each term in the numerator
  $=$= $\lim_{h\rightarrow0}(nx^{n-1}+...+nh^{n-2}x+h^{n-1})$limh0(nxn1+...+nhn2x+hn1) Divide by $h$h
  $=$= $nx^{n-1}$nxn1 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 $a^n-b^n$anbn factorises as:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+a^{n-4}b^3+......+ab^{n-2}+b^{n-1})$anbn=(ab)(an1+an2b+an3b2+an4b3+......+abn2+bn1)

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

$(x+h)^n-x^n$(x+h)nxn $=$= $[(x+h)-x][(x+h)^{n-1}+x(x+h)^{n-2}+x^2(x+h)^{n-3}+x^3(x+h)^{n-4}+......+x^{n-1}]$[(x+h)x][(x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+......+xn1]  
  $=$= $h[(x+h)^{n-1}+x(x+h)^{n-2}+x^2(x+h)^{n-3}+x^3(x+h)^{n-4}+......+x^{n-1}]$h[(x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+......+xn1] Simplifying the first bracket

Hence,

$f'(x)$f(x) $=$= $\lim_{h\rightarrow0}\frac{(x+h)^n-x^n}{h}$limh0(x+h)nxnh  
  $=$=

$\lim_{h\to0}\left(\frac{h\left[\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right]}{h}\right)$limh0(h[(x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+...xn1]h)

 
  $=$=

$\lim_{h\to0}\left(\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right)$limh0((x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+...xn1)

Divide by $h$h
  $=$=

$x^{n-1}+x\left(x\right)^{n-2}+x^2\left(x\right)^{n-3}+x^3\left(x\right)^{n-4}+...x^{n-1}$xn1+x(x)n2+x2(x)n3+x3(x)n4+...xn1

Take the limit by letting $h=0$h=0
  $=$= $x^{n-1}+x^{n-2+1}+x^{n-3+2}+x^{n-4+3}+...x^{n-1}$xn1+xn2+1+xn3+2+xn4+3+...xn1 Rewriting powers using index laws
  $=$=

$x^{n-1}+x^{n-1}+x^{n-1}+x^{n-1}+...x^{n-1}$xn1+xn1+xn1+xn1+...xn1

Simplify powers
  $=$= $nx^{n-1}$nxn1 Each power was the same and there were $n$n terms

 

Practice questions

Question 1

Find the gradient of $f\left(x\right)=x^4$f(x)=x4 at $x=2$x=2.

Denote this gradient by $f'\left(2\right)$f(2).

Question 2

Consider the function $y=\frac{1}{x^2}$y=1x2.

  1. Rewrite the function in negative index form.

  2. Determine the derivative of $y=\frac{1}{x^2}$y=1x2.

Question 3

Determine the derivative of $y=x^{\frac{6}{5}}$y=x65.

QUESTION 4

Use the applet below to explore how the gradient of the tangent changes at different points along $y=x^2$y=x2. Then answer the questions that follow.

  1. Which feature of the gradient function tells us whether $y=x^2$y=x2 is increasing or decreasing?

    The gradient function is decreasing when $y=x^2$y=x2 is increasing, and increasing when $y=x^2$y=x2 is decreasing.

    A

    The gradient function is increasing when $y=x^2$y=x2 is increasing, and decreasing when $y=x^2$y=x2 is decreasing.

    B

    The gradient function is negative when $y=x^2$y=x2 is increasing, and positive when $y=x^2$y=x2 is decreasing.

    C

    The gradient function is positive when $y=x^2$y=x2 is increasing, and negative when $y=x^2$y=x2 is decreasing.

    D
  2. For $x>0$x>0, is the gradient of the tangent positive or negative?

    Positive

    A

    Negative

    B
  3. For $x\ge0$x0, as the value of $x$x increases how does the gradient of the tangent line change?

    The gradient of the tangent line increases at a constant rate.

    A

    The gradient of the tangent line increases at an increasing rate.

    B

    The gradient of the tangent line remains constant.

    C
  4. For $x<0$x<0, is the gradient of the tangent positive or negative?

    Positive

    A

    Negative

    B
  5. For $x<0$x<0, as the value of $x$x increases how does the gradient of the tangent line change?

    The gradient of the tangent line remains constant.

    A

    The gradient of the tangent line increases at a constant rate.

    B

    The gradient of the tangent line increases at an increasing rate.

    C
  6. Complete the following statement:

    "For $y=x^2$y=x2, the gradient of the tangent line changes at a constant rate. This means the derivative $y'$y is a             function."

    cubic

    A

    linear

    B

    constant

    C

    quadratic

    D

Outcomes

2.4.2.1

examine examples of variable rates of change of non-linear functions

2.4.2.2

establish the formula 𝑑/𝑑x (𝑥^𝑛)=𝑛𝑥^(𝑛−1) for positive integers

2.4.2.5

calculate derivatives of power and polynomial functions

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