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 $\delta\rightarrow0$δ→0.
We write
$f'(x)=\lim_{\delta\rightarrow0}\frac{f(x+\delta)-f(x)}{\delta}$f′(x)=limδ→0f(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 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.
Quadratic terms
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)=limh→0(x+h)2−x2h.
(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 |
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.
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.
The gradient function is increasing when $y=x^2$y=x2 is increasing, and decreasing when $y=x^2$y=x2 is decreasing.
The gradient function is negative when $y=x^2$y=x2 is increasing, and positive when $y=x^2$y=x2 is decreasing.
The gradient function is positive when $y=x^2$y=x2 is increasing, and negative when $y=x^2$y=x2 is decreasing.
The gradient function is decreasing when $y=x^2$y=x2 is increasing, and increasing when $y=x^2$y=x2 is decreasing.
The gradient function is increasing when $y=x^2$y=x2 is increasing, and decreasing when $y=x^2$y=x2 is decreasing.
The gradient function is negative when $y=x^2$y=x2 is increasing, and positive when $y=x^2$y=x2 is decreasing.
The gradient function is positive when $y=x^2$y=x2 is increasing, and negative when $y=x^2$y=x2 is decreasing.
For $x>0$x>0, is the gradient of the tangent positive or negative?
Positive
Negative
Positive
Negative
For $x\ge0$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.
The gradient of the tangent line increases at an increasing rate.
The gradient of the tangent line remains constant.
The gradient of the tangent line increases at a constant rate.
The gradient of the tangent line increases at an increasing rate.
The gradient of the tangent line remains constant.
For $x<0$x<0, is the gradient of the tangent positive or negative?
Positive
Negative
Positive
Negative
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.
The gradient of the tangent line increases at a constant rate.
The gradient of the tangent line increases at an increasing rate.
The gradient of the tangent line remains constant.
The gradient of the tangent line increases at a constant rate.
The gradient of the tangent line increases at an increasing rate.
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
linear
constant
quadratic
cubic
linear
constant
quadratic
Consider the functions $f\left(x\right)=x^5$f(x)=x5 and $g\left(x\right)=x^4$g(x)=x4.
Which of the following shows the graph of $f\left(x\right)$f(x) and its derivative?
Which of the following shows the graph of $g\left(x\right)$g(x) and its derivative?
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.
Near the origin, the derivative has a greater value than the function.
A function and its derivative have the same sign for all values of $x$x.
If the degree of a function is even, then the degree of its derivative is odd and vice versa.
The graph of the derivative can be found by translating and/or stretching the original function.
Near the origin, the derivative has a greater value than the function.
A function and its derivative have the same sign for all values of $x$x.
If the degree of a function is even, then the degree of its derivative is odd and vice versa.
Consider the function $f\left(x\right)=x^2$f(x)=x2.
Using first principles, find the derivative of $f\left(x\right)=x^2$f(x)=x2.
Show all steps of working.
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 | $x^2$x2 | $x^3$x3 | $x^4$x4 | $x^5$x5 |
---|---|---|---|---|
$y'$y′ | $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.
Deduce the general form of the derivative of $y=x^n$y=xn for any value of $n$n.
Apply differentiation and anti-differentiation techniques to polynomials
Apply calculus methods in solving problems