We have seen how to find the derivative of functions of the form $f(x)=x^n$f(x)=xn using the power rule but what about multiples or combinations of these functions such as: $f(x)=5x^2$f(x)=5x2, $g(x)=3x^3+5x^2-6$g(x)=3x3+5x2−6, $h(x)=\frac{4}{x}+\sqrt{x}$h(x)=4x+√x. If we returned to first principles and used the linear properties of limits we would discover the following useful properties for derivatives.
Power Rule:
For a function $f(x)=x^n$f(x)=xn, $f'(x)=nx^{n-1}$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 $f(x)=kg(x)$f(x)=kg(x), where $k$k is a constant, then $f'(x)=kg'(x)$f′(x)=kg′(x)
In particular, if $f(x)=ax^n$f(x)=axn, then $f'(x)=nax^{n-1}$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 $f(x)=g(x)\pm h(x)$f(x)=g(x)±h(x) then $f'(x)=g'(x)\pm h'(x)$f′(x)=g′(x)±h′(x)
This means we can differentiate each individual term.
Worked examples
Example 1
Find the derivatives of the following functions.
Think: These are all multiples of power functions, hence use $f'(x)=anx^{n-1}$f′(x)=anxn−1.
a) $f(x)=3x^2$f(x)=3x2
| $f'(x)$f′(x) | $=$= | $3\times2x$3×2x |
| $=$= | $6x$6x |
b) $g(m)=2m^4$g(m)=2m4
| $g'(m)$g′(m) | $=$= | $2\times4m^3$2×4m3 |
| $=$= | $8m^3$8m3 |
c) $h(t)=-3t^{\frac{3}{2}}$h(t)=−3t32
| $h'(t)$h′(t) | $=$= | $-3\times\frac{3}{2}t^{\frac{3}{2}-1}$−3×32t32−1 |
| $=$= | $-\frac{9}{2}t^{\frac{1}{2}}$−92t12 |
d) $g(x)=\frac{3}{5x^2}$g(x)=35x2, writing your answer with positive indices.
Firstly we need the function in power form, converting we obtain $g(x)=\frac{3}{5}x^{-2}$g(x)=35x−2.
| $g'(x)$g′(x) | $=$= | $\frac{3}{5}\times\left(-2x^{-2-1}\right)$35×(−2x−2−1) | Careful with negative powers |
| $=$= | $-\frac{6}{5}x^{-3}$−65x−3 | ||
| $=$= | $-\frac{6}{5x^3}$−65x3 | Equivalent form with a positive power |
Example 2
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) $f(x)=4x^2+3x+2$f(x)=4x2+3x+2
Thus, $f'(x)=8x+3$f′(x)=8x+3 (remember that the derivative of a constant term is $0$0)
b) $f(x)=3x^3-3x^2$f(x)=3x3−3x2
Thus, $f'(x)=9x^2-6x$f′(x)=9x2−6x
c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x−3−2x+√x
Firstly we need to turn the $\sqrt{x}$√x into a power. We have$f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x−3−2x+x12.
Thus, the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$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.
$\frac{d}{dx}\left[f(x)\times g(x)\right]\ne f'(x)\times g'(x)$ddx[f(x)×g(x)]≠f′(x)×g′(x) There are special rules for products and quotients you will use in Year 12. For now when you come across a product of two functions 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 you can split a fraction into individual terms with the same denominator.
Worked examples
Example 3
Find the derivative of the following functions.
a) $f(x)=(x+3)(x-1)$f(x)=(x+3)(x−1)
First expand:
| $f(x)$f(x) | $=$= | $x^2+3x-x-3$x2+3x−x−3 |
| $=$= | $x^2+2x-3$x2+2x−3 |
Then differentiate: $f'(x)=2x+2$f′(x)=2x+2
b) $g(x)=x(x+1)^2$g(x)=x(x+1)2
First expand:
| $g(x)$g(x) | $=$= | $x(x^2+2x+1)$x(x2+2x+1) |
| $=$= | $x^3+2x^2+x$x3+2x2+x |
Then differentiate: $g'(x)=3x^2+4x+1$g′(x)=3x2+4x+1
Example 4
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:
| $f(x)$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, $f'(x)=\frac{1}{2}-4x^{-2}$f′(x)=12−4x−2
Example 5
For the function $f(x)=10\sqrt{x}$f(x)=10√x:
a) Evaluate $f'(4)$f′(4).
Think: Find the function $f'(x)$f′(x) and the evaluate the function at $x=4$x=4.
Do: The function can be rewritten as $f(x)=10x^{\frac{1}{2}}$f(x)=10x12.
Hence,
| $f'(x)$f′(x) | $=$= | $10\times\frac{1}{2}x^{-\frac{1}{2}}$10×12x−12 |
| $=$= | $\frac{5}{\sqrt{x}}$5√x |
Thus,
| $f'(4)$f′(4) | $=$= | $\frac{5}{\sqrt{4}}$5√4 |
| $=$= | $\frac{5}{2}$52 |
b) Find the $x$x-coordinate where the gradient of the tangent to the function is equal to $1$1.
Think: We need to solve the equation $f'(x)=1$f′(x)=1 for $x$x.
Do:
| $\frac{5}{\sqrt{x}}$5√x | $=$= | $1$1 |
| $\sqrt{x}$√x | $=$= | $5$5 |
| $\therefore x$∴x | $=$= | $25$25 |
Practice questions
question 1
Differentiate $y=2x^3-3x^2-4x+13$y=2x3−3x2−4x+13.
question 2
Differentiate $y=7ax^7-2bx^3$y=7ax7−2bx3, where $a$a and $b$b are constants.
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
question 4
Consider the function $f\left(x\right)=\left(\sqrt{x}+10x^2\right)^2$f(x)=(√x+10x2)2
Express the function $f\left(x\right)$f(x) in expanded form, with all terms written as powers of $x$x.
Hence find the derivative $f'\left(x\right)$f′(x) of the function $f\left(x\right)=\left(\sqrt{x}+10x^2\right)^2$f(x)=(√x+10x2)2
Remove all fractional indices in your final answer.
question 5
Find the derivative of $y=\frac{x+1}{\sqrt[7]{x}}$y=x+17√x.
Give your answer with positive indices.
question 6
Consider the function $f\left(x\right)=x^3-4x$f(x)=x3−4x.
Determine $f'\left(x\right)$f′(x).
Determine $f'\left(4\right)$f′(4).
Determine $f'\left(-4\right)$f′(−4).
Determine the $x$x-coordinates of the points at which $f'\left(x\right)=71$f′(x)=71.
State your solutions on the same line, separated by a comma.