Functions of the type $f(x)=x^n$f(x)=xn are called power functions. From the last lesson on first principles, you may have noticed some patterns when calculating the derivatives. 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 $n-1$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:
| $f(x)$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, it makes sense to think 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 this formula is called 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 established, the power rule can be used as a shortcut to find the derivative without having to use first principles every time.
For a function $f(x)=x^n$f(x)=xn, the derivative is $f'(x)=nx^{n-1}$f′(x)=nxn−1, for $n$n any real number.
Let's try applying this rule, and then explore 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}$32t32−1 |
| $=$= | $\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=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}$52x52−1 |
| $=$= | $\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
First principles can be used to prove the power rule for positive integers $n$n. To prove the more general case requires knowledge for differentiation of exponential functions which will not be encountered until 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, there are two choices to move forward, either to expand $\left(x+h\right)^n-x^n$(x+h)n−xn or to factorise it. In fact, both cases will give the same result.
Method 1 - Expansion
When we expand $\left(x+h\right)^n-x^n$(x+h)n−xn it gives 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 the final result. So, the shorthand version can be used:
| $f'(x)$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 a $h$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 $n$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.
Rewrite the function in negative index form.
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.
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.
AThe gradient function is increasing when $y=x^2$y=x2 is increasing, and decreasing when $y=x^2$y=x2 is decreasing.
BThe gradient function is negative when $y=x^2$y=x2 is increasing, and positive when $y=x^2$y=x2 is decreasing.
CThe gradient function is positive when $y=x^2$y=x2 is increasing, and negative when $y=x^2$y=x2 is decreasing.
DFor $x>0$x>0, is the gradient of the tangent positive or negative?
Positive
ANegative
BFor $x\ge0$x≥0, 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.
AThe gradient of the tangent line increases at an increasing rate.
BThe gradient of the tangent line remains constant.
CFor $x<0$x<0, is the gradient of the tangent positive or negative?
Positive
ANegative
BFor $x<0$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.
AThe gradient of the tangent line increases at a constant rate.
BThe gradient of the tangent line increases at an increasing rate.
CComplete 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
Alinear
Bconstant
Cquadratic
D