Recall from Year 11 that the derivative of a function gives the gradient of the function at a particular $x$x value. We can find the derivative of functions of the form $f\left(x\right)=x^n$f(x)=xn using the power rule and for multiples or combinations of these functions such as: $f\left(x\right)=5x^2$f(x)=5x2, $g\left(x\right)=3x^3+5x^2-6$g(x)=3x3+5x2−6, $h\left(x\right)=\frac{4}{x}+\sqrt{x}$h(x)=4x+√x, we can apply the power rule together with the following properties of derivatives:
Power Rule:
For a function $f\left(x\right)=x^n$f(x)=xn, $f'\left(x\right)=nx^{n-1}$f′(x)=nxn−1, for $n$n any real number.
Derivative of a constant multiple:
Derivative of constant multiple of a function is the multiple of the derivative. That is if $f\left(x\right)=kg\left(x\right)$f(x)=kg(x), where $k$k is a constant, then $f'\left(x\right)=kg'\left(x\right)$f′(x)=kg′(x)
In particular, if $f\left(x\right)=ax^n$f(x)=axn, then $f'\left(x\right)=nax^{n-1}$f′(x)=naxn−1, where $a$a is a constant.
Derivative of a sum or difference:
Derivative of sum is equal to the sum of the derivatives. That is if $f\left(x\right)=g\left(x\right)\pm h\left(x\right)$f(x)=g(x)±h(x) then $f'\left(x\right)=g'\left(x\right)\pm h'\left(x\right)$f′(x)=g′(x)±h′(x)
This means we can differentiate each term individually.
Worked examples
Example 1
Find the derivative of the following functions.
Think: These functions can be considered as the sum of the function for each individual term. So we can differentiate the whole by finding the derivative of each part using the power and constant multiple rules.
(a) $f\left(x\right)=4x^2+3x+2$f(x)=4x2+3x+2
Thus, $f'\left(x\right)=8x+3$f′(x)=8x+3 (remember that the derivative of a constant term is $0$0)
(b) $f\left(x\right)=3x^3-3x^2$f(x)=3x3−3x2
Thus, $f'\left(x\right)=9x^2-6x$f′(x)=9x2−6x
(c) $f\left(x\right)=6x^{-3}-2x+\sqrt{x}$f(x)=6x−3−2x+√x
Firstly we need to turn the $\sqrt{x}$√x into a power. We have$f\left(x\right)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x−3−2x+x12.
Thus, the derivative $f'\left(x\right)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f′(x)=−18x−4−2+12x−12
Our rules encountered so far do not tell us how to differentiate in the case where a function is the product of two functions or the quotient of two functions.
$\frac{d}{dx}\left[f\left(x\right)\times g\left(x\right)\right]\ne f'\left(x\right)\times g'\left(x\right)$ddx[f(x)×g(x)]≠f′(x)×g′(x) There are special rules for products and quotients we will look at in the next lessons. For now when we come across a product of two functions, we will expand to produce a function with all terms of the form $ax^n$axn. For a quotient perform the division and simplify using index laws. Remember we can split a fraction into individual terms with the same denominator.
Worked examples
Example 2
Find the derivative of the following function $g\left(x\right)=x\left(x+1\right)^2$g(x)=x(x+1)2.
First expand:
| $g\left(x\right)$g(x) | $=$= | $x\left(x^2+2x+1\right)$x(x2+2x+1) |
| $=$= | $x^3+2x^2+x$x3+2x2+x |
Then differentiate:
$g'\left(x\right)=3x^2+4x+1$g′(x)=3x2+4x+1
Example 3
Differentiate $f\left(x\right)=\frac{x^2-4x+8}{2x}$f(x)=x2−4x+82x .
Think: The fraction bar (vinculum) acts like brackets, so we need to divide every term in the numerator by the denominator. We can split the fraction into three terms and then use index laws to simplify each term.
Do:
| $f\left(x\right)$f(x) | $=$= | $\frac{x^2-4x+8}{2x}$x2−4x+82x |
| $=$= | $\frac{x^2}{2x}-\frac{4x}{2x}+\frac{8}{2x}$x22x−4x2x+82x | |
| $=$= | $\frac{x}{2}-2+\frac{4}{x}$x2−2+4x | |
| $=$= | $\frac{1}{2}x-2+4x^{-1}$12x−2+4x−1 |
We now have each term in a form that we can apply the power rule.
Hence, $f'\left(x\right)=\frac{1}{2}-4x^{-2}$f′(x)=12−4x−2
Practice questions
question 1
Differentiate $y=2x^3-3x^2-4x+13$y=2x3−3x2−4x+13.
question 2
Consider the function $y=\frac{5x\sqrt{x}}{4x^5}$y=5x√x4x5.
Fully simplify the function, expressing your answer with a negative index.
Find $\frac{dy}{dx}$dydx.
question 3
Consider the function $y=\left(x+4\right)^2$y=(x+4)2
Express the function $y$y in expanded form.
Hence find the derivative $\frac{dy}{dx}$dydx of the function $y=\left(x+4\right)^2$y=(x+4)2
Applications
Before we look at further real-life application problems, the following application-style problems are common and involve using and manipulating the gradient function in one of the following ways:
- Determine the equation of the tangent at a given point on the curve
- Determine the gradient at a given point on the curve
- Determine the point(s) on a curve where a given gradient exists
- Using the information about gradients and the original function to determine unknown coefficients in the polynomial $f\left(x\right)$f(x).
Equations of tangents
A tangent to a function is a straight line and as such we can use our knowledge of linear functions to find the equation of a tangent. Our new technique of differentiation will allow us to find the gradient of the tangent to a function at any given point.
For a function $y=f\left(x\right)$y=f(x) the equation of the tangent at the point of contact $\left(x_1,y_1\right)$(x1,y1) can be found using either:
- $y=mx+c$y=mx+c (gradient-intercept form)
- $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1) (point-gradient formula)
Where the gradient of the tangent is $m=f'\left(x_1\right)$m=f′(x1).
Finding the equation of a tangent line from a graph
From a graph look for two easily identifiable points then calculate the gradient using $m=\frac{rise}{run}$m=riserun. Then use the gradient and one of the points and substitute into one of the forms above to find the equation. If it is clear on a graph a convenient point to use would be the $y$y-intercept.
Finding the equation of the tangent to $y=f\left(x\right)$y=f(x) at $x=a$x=a
Steps: The equation is of the form $y=mx+c$y=mx+c, we need to find $m$m and then $c$c.
- Find the gradient, $m$m, by evaluating $m=f'\left(a\right)$m=f′(a)
- Find the point of contact, the shared point between the function and the tangent by evaluating $y=f\left(a\right)$y=f(a). Giving us the point of contact as $\left(a,f\left(a\right)\right)$(a,f(a)). (If the point is given in the question, simply state it.)
- Find $c$c by substituting the point of contact and gradient into the equation $y=mx+c$y=mx+c and then rearrange. (Or use the point gradient form of the equation)
- State the equation of the tangent.
Finding the equation of the normal to the curve
The normal line is perpendicular to the tangent line. Therefore the gradient of the normal line is the negative reciprocal of the gradient of the tangent. We find the equation of the normal to a curve in exactly the same way as the tangent to the curve.
Worked examples
Example 4
Find the equation of the tangent to $f\left(x\right)$f(x) at the point pictured below.
Think: We can see the point of contact and the $y$y-intercept clearly. Use these two points to find the gradient and then write the equation in the form $y=mx+c$y=mx+c.
Do: The point of contact is $\left(1,-3\right)$(1,−3) and the $y$y-intercept is $\left(0,-5\right)$(0,−5). Thus, the gradient is:
| $m$m | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
| $=$= | $\frac{-3-\left(-5\right)}{1-0}$−3−(−5)1−0 | |
| $=$= | $2$2 |
Since we have the $y$y-intercept we know $c=-5$c=−5, and hence the equation of the tangent is $y=2x-5$y=2x−5.
Example 5
Find the equation of the tangent to $f\left(x\right)=\sqrt{x}$f(x)=√x at $x=4$x=4.
Think: We have not been given the point of contact but we can evaluate the function at $x=4$x=4 to find it. We also need to find the derivative to find the gradient of the tangent.
Do:
Find the gradient of tangent:
| $f\left(x\right)$f(x) | $=$= | $x^{\frac{1}{2}}$x12 |
| $f'\left(x\right)$f′(x) | $=$= | $\frac{1}{2}x^{-\frac{1}{2}}$12x−12 |
| $=$= | $\frac{1}{2\sqrt{x}}$12√x | |
| $\therefore f'\left(4\right)$∴f′(4) | $=$= | $\frac{1}{2\sqrt{4}}$12√4 |
| $=$= | $\frac{1}{4}$14 |
Find the point of contact, when $x=4$x=4:
| $f\left(4\right)$f(4) | $=$= | $\sqrt{4}$√4 |
| $=$= | $2$2 |
Thus, the point of contact is $\left(4,2\right)$(4,2).
Find $c$c:
The tangent is of the form $y=\frac{1}{4}x+c$y=14x+c and passes through $\left(4,2\right)$(4,2). Substituting into the equation we get:
| $2$2 | $=$= | $\frac{1}{4}\left(4\right)+c$14(4)+c |
| $2$2 | $=$= | $1+c$1+c |
| $\therefore c$∴c | $=$= | $1$1 |
Hence, the equation of the tangent to $f\left(x\right)$f(x) at $x=4$x=4 is $y=\frac{1}{4}x+1$y=14x+1.
Reflect: Alternatively we could use the point-gradient formula and substitute $x_1=4$x1=4, $y_1=2$y1=2 and $m=\frac{1}{4}$m=14 into the rule $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1) which gives $y-2=\frac{1}{4}\left(x-4\right)$y−2=14(x−4) and simplifies to $y=\frac{1}{4}x+1$y=14x+1
Problem solving using gradient information
Common problem solving questions include using the derivative to determine the point(s) on a curve where a given gradient occurs or using the information about gradients and the original function to determine unknown coefficients in the original function.
Worked example
Example 6
The function $f\left(x\right)=x^3+ax^2+bx+c$f(x)=x3+ax2+bx+c has a $y$y-intercept of $3$3. The function has gradient of zero at $x=1$x=1 and a root at $x=-3$x=−3. Determine the values of $a$a, $b$band $c$c.
Think: Break the information into parts and determine if the information given is about the function itself or its derivative.
- The $y$y-intercept tells us the original function passes through $\left(0,3\right)$(0,3)
- The gradient at $x=1$x=1 tell us that the derivative equals zero when $x=1$x=1, that is $f'\left(1\right)=0$f′(1)=0
- The root tells us the original function passes through $\left(-3,0\right)$(−3,0)
We have three pieces of information and three unknowns, so we should be able to solve using simultaneous equations.
Do: We can begin by using the information about the $y$y-intercept. Substituting $\left(0,3\right)$(0,3) into the original function we get:
| $0^3+a\times0^2+b\times0+c$03+a×02+b×0+c | $=$= | $3$3 |
| $\therefore c$∴c | $=$= | $3$3 |
To use the information about the gradient, we will need to first find ourselves the derivative.
$f'\left(x\right)=3x^2+2ax+b$f′(x)=3x2+2ax+b
Using the information $f'\left(1\right)=0$f′(1)=0, we obtain the equation:
| $3\left(1\right)^2+2a\left(1\right)+b$3(1)2+2a(1)+b | $=$= | $0$0 | |
| $2a+b$2a+b | $=$= | $-3$−3 | ....Equation $1$1 |
To use the information about the $x$x-intercept, we can substitute $\left(-3,0\right)$(−3,0) into the original function to obtain the equation:
| $\left(-3\right)^3+a\left(-3\right)^2+b\left(-3\right)+3$(−3)3+a(−3)2+b(−3)+3 | $=$= | $0$0 | |
| $-27+9a-3b+3$−27+9a−3b+3 | $=$= | $0$0 | |
| $9a-3b$9a−3b | $=$= | $24$24 | ....Equation $2$2 |
Solving equation $1$1 and $2$2 simultaneously (either with the elimination method, substitution method or with technology) we find that $a=1$a=1 and $b=-5$b=−5. Thus the original function was $f\left(x\right)=x^3+x^2-5x+3$f(x)=x3+x2−5x+3.
Practice questions
Question 4
Consider the function $f\left(x\right)=x^2+5x$f(x)=x2+5x.
Find the $x$x-coordinate of the point at which $f\left(x\right)$f(x) has a gradient of $13$13.
Hence state the coordinates of the point on the curve where the gradient is $13$13.
Question 5
Consider the curve $f\left(x\right)$f(x) drawn below along with $g\left(x\right)$g(x), which is a tangent to the curve.
What are the coordinates of the point at which $g\left(x\right)$g(x) is a tangent to the curve $f\left(x\right)$f(x)?
Note that this point has integer coordinates. Give your answer in the form $\left(a,b\right)$(a,b).
What is the gradient of the tangent line?
Hence determine the equation of the line $y=g\left(x\right)$y=g(x).
Question 6
Consider the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15.
Find $f'\left(x\right)$f′(x).
Find the gradient of the tangent to the curve at the point $\left(4,63\right)$(4,63).
Determine the equation of the tangent to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).
Express the equation of the tangent line in the form $y=mx+c$y=mx+c.
Find the gradient of the normal to the curve at the point $\left(4,63\right)$(4,63).
Determine the equation of the normal to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).
Question 7
The curve $y=ax^3+bx^2+2x-17$y=ax3+bx2+2x−17 has a gradient of $58$58 at the point $\left(2,31\right)$(2,31).
Use the fact that the gradient of the curve at the point $\left(2,31\right)$(2,31) is $58$58 to express $b$b in terms of $a$a.
Use the fact that the curve passes through the point $\left(2,31\right)$(2,31) to express $b$b in terms of $a$a.
Hence solve for $a$a.
Hence solve for $b$b.