NZ Level 7 (NZC) Level 2 (NCEA)
Deriving the Power Rule
Lesson

To differentiate a function means to derive a related function that gives the gradient of the original function at each point of its domain.

We visualise the process for doing this by considering a secant line that cuts a graph at a point $x$x and also at a nearby point $x+\delta$x+δ. We form an expression for its gradient. We then imagine $\delta$δ becoming progressively smaller so that eventually, in the limit, the secant becomes a tangent. The gradient of the tangent is what we require. The process is illustrated in the following diagram.

The expression for the gradient of the secant line is the familiar 'rise-over-run'. Suppose we are differentiating a function $f$f. Then, the gradient of the secant passing through $x$x and $x+\delta$x+δ is $\frac{f(x+\delta)-f(x)}{(x+\delta)-x}$f(x+δ)f(x)(x+δ)x. We simplify this and find the limit as $\delta\rightarrow0$δ0.

We write

$f'(x)=\lim_{\delta\rightarrow0}\frac{f(x+\delta)-f(x)}{\delta}$f(x)=limδ0f(x+δ)f(x)δ.

This limit is called the derivative of the function $f$f and the process is called differentiation. The function $f'$f is also called the gradient function for $f$f.

## Linear functions

Linear functions are of the form $f(x)=ax+b$f(x)=ax+b where a and b are constants. The variable $x$x is raised to the power $1$1. We already know that the graph of such a function has a constant gradient of $a$a so, we can use it to test our limit statement. We have

 $f'(x)$f′(x) $=$= $\lim_{\delta\rightarrow0}\frac{a(x+\delta)+b-(ax+b)}{\delta}$limδ→0a(x+δ)+b−(ax+b)δ​ $=$= $\lim_{\delta\rightarrow0}\frac{a\delta}{\delta}$limδ→0aδδ​ $=$= $a$a

We were able to cancel out the $\delta$δ from numerator and denominator because $\delta$δ is not yet $0$0 until the limit is reached.

Thus, the procedure has produced the expected result, $f'(x)=a$f(x)=a for all values of $x$x.

##### Example 2

Consider the function $g(x)=x^2$g(x)=x2. According to the procedure, we have

$g'(x)=\lim_{h\rightarrow0}\frac{(x+h)^2-x^2}{h}$g(x)=limh0(x+h)2x2h.

(We have used the symbol $h$h instead of $\delta$δ, for no particular reason.)

Then,

 $g'(x)$g′(x) $=$= $\lim_{h\rightarrow0}\frac{x^2+2xh+h^2-x^2}{h}$limh→0x2+2xh+h2−x2h​ $=$= $\lim_{h\rightarrow0}\frac{2xh+h^2}{h}$limh→02xh+h2h​ $=$= $\lim_{h\rightarrow0}\frac{h(2x+h)}{h}$limh→0h(2x+h)h​ $=$= $\lim_{h\rightarrow0}\frac{2x+h}{1}$limh→02x+h1​ $=$= $2x$2x

##### Example 3

Positive integer power terms

If we look at what happens with a more general function $k(x)=x^n$k(x)=xn where $n$n is a positive integer, we should be able to find a method for finding the derivative of any term of this kind.

We write

 $k'(x)$k′(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​ $=$= $\lim_{h\rightarrow0}\frac{h(nx^{n-1}+...nh^{n-2}x+h^{n-1})}{h}$limh→0h(nxn−1+...nhn−2x+hn−1)h​ $=$= $\lim_{h\rightarrow0}(nx^{n-1}+...+nh^{n-2}x+h^{n-1})$limh→0(nxn−1+...+nhn−2x+hn−1) $=$= $nx^{n-1}$nxn−1

We omitted the middle terms in the binomial expansion. In the limit, most of the terms disappeared because they had $h$h as a factor.

With a little more work and a more general form of the binomial expansion, we can show that the differentiation of a power follows this rule for every exponent $n$n, positive, negative or zero, integer or not.

#### Practice Questions

##### QUESTION 1

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

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

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

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

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

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

D

cubic

A

linear

B

constant

C

D

##### QUESTION 2

Consider the functions $f\left(x\right)=x^5$f(x)=x5 and $g\left(x\right)=x^4$g(x)=x4.

1. Which of the following shows the graph of $f\left(x\right)$f(x) and its derivative?

A

B

C

D

A

B

C

D
2. Which of the following shows the graph of $g\left(x\right)$g(x) and its derivative?

A

B

C

D

A

B

C

D
3. Which of the following statements are true? Select all that apply.

The graph of the derivative can be found by translating and/or stretching the original function.

A

Near the origin, the derivative has a greater value than the function.

B

A function and its derivative have the same sign for all values of $x$x.

C

If the degree of a function is even, then the degree of its derivative is odd and vice versa.

D

The graph of the derivative can be found by translating and/or stretching the original function.

A

Near the origin, the derivative has a greater value than the function.

B

A function and its derivative have the same sign for all values of $x$x.

C

If the degree of a function is even, then the degree of its derivative is odd and vice versa.

D

##### QUESTION 3

Consider the function $f\left(x\right)=x^2$f(x)=x2.

1. Using first principles, find the derivative of $f\left(x\right)=x^2$f(x)=x2.

Show all steps of working.

2. The derivative of $y=x^n$y=xn for other values of $n$n can be found using first principles in a similar way. The results for $n=2,3,4$n=2,3,4 and $5$5 are summarised in the table below.

 $y$y $y'$y′ $x^2$x2 $x^3$x3 $x^4$x4 $x^5$x5 $2x$2x $3x^2$3x2 $4x^3$4x3 $5x^4$5x4

By observing the pattern in the table, deduce the derivative of the function $g\left(x\right)=x^8$g(x)=x8.

3. Deduce the general form of the derivative of $y=x^n$y=xn for any value of $n$n.

### Outcomes

#### M7-10

Apply differentiation and anti-differentiation techniques to polynomials

#### 91262

Apply calculus methods in solving problems