Functions and algebra: Investigate and use instantaneous rate of change
Unit 3: Find the derivative using rules for differentiation
Natashia Bearam-Edmunds
Unit outcomes
By the end of this unit you will be able to:
- Apply the constant, sum and difference rules for differentiation.
- Use notation appropriately.
What you should know
Before you start this unit, make sure you can:
- Find the derivative by using first principles as shown in unit 2 of this subject outcome.
Introduction
It can be quite a long and challenging process to find derivatives of functions by using the definition of the derivative or first principles. For example, when we previously found the derivative by first principles of [latex]\scriptsize k(x)=-\displaystyle \frac{2}{x}[/latex] (unit 2 Exercise 2.2) we saw that we had to work with complicated fractions before evaluating the limit.
In this unit, we will learn how to find the derivative of a function with rules instead of with first principles. Unless a question specifically requires the use of the definition to find the derivative, we can always use the rules to derive.
The basics
The functions [latex]\scriptsize f(x)=c[/latex] and [latex]\scriptsize g(x)={{x}^{n}}[/latex], where [latex]\scriptsize n[/latex] is a positive integer, are the building blocks from which all polynomials and rational functions are created. To find derivatives of polynomials and rational functions without using the definition of the derivative, we must first develop formulae for differentiating these two basic functions.
Activity 3.1: Discover the basic rules for differentiation
Time required: 35 minutes
What you need:
- pen and paper
What to do
- Differentiate the following by first principles:
- [latex]\scriptsize f(x)=2[/latex]
- [latex]\scriptsize f(x)=6[/latex]
- [latex]\scriptsize f(x)=x[/latex]
- [latex]\scriptsize f(x)=-2x[/latex]
- [latex]\scriptsize f(x)={{x}^{2}}[/latex]
- [latex]\scriptsize f(x)=4{{x}^{2}}[/latex]
- Complete the table below:
[latex]\scriptsize f(x)[/latex] [latex]\scriptsize {f}'(x)[/latex] Power of [latex]\scriptsize x[/latex] in [latex]\scriptsize f(x)[/latex] Co-efficient of [latex]\scriptsize {f}'(x)[/latex] Power of [latex]\scriptsize x[/latex] in [latex]\scriptsize {f}'(x)[/latex] [latex]\scriptsize 2[/latex] [latex]\scriptsize 6[/latex] [latex]\scriptsize x[/latex] [latex]\scriptsize -2x[/latex] [latex]\scriptsize {{x}^{2}}[/latex] [latex]\scriptsize 4{{x}^{2}}[/latex] - Can you see a pattern for finding the derivative of a constant function [latex]\scriptsize c[/latex]?
- Can you see a pattern for finding the derivative of a power function [latex]\scriptsize {{x}^{n}}[/latex]?
What did you find?
- .
- [latex]\scriptsize f(x)=2[/latex]
For this function, both [latex]\scriptsize \displaystyle f\left( x \right)=2[/latex] and [latex]\scriptsize \displaystyle f\left( {x+h} \right)=2[/latex], so we get:
[latex]\scriptsize \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{2-2}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{0}{h}\\&=0\end{align*}[/latex] - [latex]\scriptsize f(x)=6[/latex]
For this function, both [latex]\scriptsize \displaystyle f\left( x \right)=6[/latex] and [latex]\scriptsize \displaystyle f\left( {x+h} \right)=6[/latex], so we get:
[latex]\scriptsize \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{6-6}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{0}{h}\\&=0\end{align*}[/latex] - [latex]\scriptsize f(x)=x[/latex]
For this function, [latex]\scriptsize \displaystyle f\left( x \right)=x[/latex] and [latex]\scriptsize \displaystyle f\left( {x+h} \right)=x+h[/latex], so we get:
[latex]\scriptsize \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{(x+h)-(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{h}{h}\\&=1\end{align*}[/latex] - [latex]\scriptsize f(x)=-2x[/latex]
For this function, [latex]\scriptsize \displaystyle f\left( x \right)=-2x[/latex] and
[latex]\scriptsize \displaystyle \begin{align*}f\left( {x+h} \right)&=-2(x+h)\\&=-2x-2h\end{align*}[/latex]
so we get:
[latex]\scriptsize \displaystyle \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{-2(x+h)-(-2x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{-2x-2h+2x}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{-2h}}{h}\\&=-2\end{align*}[/latex] - [latex]\scriptsize f(x)={{x}^{2}}[/latex]
For this function, [latex]\scriptsize \displaystyle f\left( x \right)={{x}^{2}}[/latex] and
[latex]\scriptsize \displaystyle \begin{align*}f\left( {x+h} \right)&={{(x+h)}^{2}}\\&={{x}^{2}}+2xh+{{h}^{2}}\end{align*}[/latex]
so we get:
[latex]\scriptsize \displaystyle \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{{{x}^{2}}+2xh+{{h}^{2}}-{{x}^{2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{2xh+{{h}^{2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{h(2x+h)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,(2x+h)\\&=2x\end{align*}[/latex] - [latex]\scriptsize f(x)=4{{x}^{2}}[/latex]
For this function, [latex]\scriptsize \displaystyle f\left( x \right)=4{{x}^{2}}[/latex] and
[latex]\scriptsize \displaystyle \begin{align*}f\left( {x+h} \right)&=4{{(x+h)}^{2}}\\&=4{{x}^{2}}+8xh+4{{h}^{2}}\end{align*}[/latex]
so we get:
[latex]\scriptsize \displaystyle \begin{align*}{f}'(x)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{4{{x}^{2}}+8xh+4{{h}^{2}}-4{{x}^{2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{8xh+4{{h}^{2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{h(8x+4h)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,(8x+4h)\\&=8x\end{align*}[/latex]
- [latex]\scriptsize f(x)=2[/latex]
- .
[latex]\scriptsize f(x)[/latex] [latex]\scriptsize {f}'(x)[/latex] Power of [latex]\scriptsize x[/latex] in [latex]\scriptsize f(x)[/latex] Co-efficient of [latex]\scriptsize {f}'(x)[/latex] Power of [latex]\scriptsize x[/latex] in [latex]\scriptsize {f}'(x)[/latex] [latex]\scriptsize 2[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize 2\times 0[/latex] None [latex]\scriptsize 6[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize 6\times 0[/latex] None [latex]\scriptsize x[/latex] [latex]\scriptsize 1[/latex] [latex]\scriptsize 1[/latex] [latex]\scriptsize 1\times 1[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize -2x[/latex] [latex]\scriptsize -2[/latex] [latex]\scriptsize 1[/latex] [latex]\scriptsize -2\times 1[/latex] [latex]\scriptsize 0[/latex] [latex]\scriptsize {{x}^{2}}[/latex] [latex]\scriptsize 2x[/latex] [latex]\scriptsize 2[/latex] [latex]\scriptsize 2\times 1[/latex] [latex]\scriptsize 1[/latex] [latex]\scriptsize 4{{x}^{2}}[/latex] [latex]\scriptsize 8x[/latex] [latex]\scriptsize 2[/latex] [latex]\scriptsize 2\times 4[/latex] [latex]\scriptsize 1[/latex] - From a) and b) we see that the derivative of a number (constant) is equal to [latex]\scriptsize 0[/latex].
- From c) to e) we can see that when we derive a power function, the exponent is multiplied by the co-efficient of the original function and the derivative’s power is one less than the original function’s power.
For example, when we found the derivative of [latex]\scriptsize f(x)=4{{x}^{2}}[/latex] we got [latex]\scriptsize \displaystyle {f}'(x)=8x[/latex]. The power of [latex]\scriptsize x[/latex] in [latex]\scriptsize f(x)[/latex] is [latex]\scriptsize 2[/latex] and the coefficient is [latex]\scriptsize 4[/latex]. The power of [latex]\scriptsize x[/latex] in [latex]\scriptsize f'(x)[/latex] is [latex]\scriptsize 1[/latex], which is one less than the power of [latex]\scriptsize f(x)[/latex]. The coefficient of [latex]\scriptsize x[/latex] in [latex]\scriptsize f'(x)[/latex] is [latex]\scriptsize 8[/latex], which is the power of [latex]\scriptsize 2[/latex] from [latex]\scriptsize f(x)[/latex] multiplied by the coefficient of [latex]\scriptsize 4[/latex].
Although it is often unwise to draw general conclusions from specific examples, Activity 3.1 is a useful way to get an idea of the patterns that are forming. We note from the activity that when we differentiate [latex]\scriptsize f(x)=x,\text{ }f(x)=-2x,\text{ }f(x)={{x}^{2}}[/latex] and [latex]\scriptsize f(x)=4{{x}^{2}}[/latex], the power of [latex]\scriptsize x[/latex] is multiplied by the coefficient of [latex]\scriptsize f(x)[/latex] in the derivative. And, the power of [latex]\scriptsize x[/latex] in the derivative decreases by [latex]\scriptsize 1[/latex].
We also saw that when we derive a constant we end up with an answer of zero.
The following two rules are the basic rules for differentiation that you will start off with.
The constant rule:
If [latex]\scriptsize f(x)=c[/latex] where [latex]\scriptsize c[/latex] is any constant then [latex]\scriptsize {f}'(x)=0[/latex].
The power rule:
If [latex]\scriptsize f(x)={{x}^{n}}[/latex] then [latex]\scriptsize f'(x)=n{{x}^{{n-1}}}[/latex] where [latex]\scriptsize n\in \mathbb{R},\text{ n}\ne \text{0}[/latex]
Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract or multiply by a constant.
The derivative of a constant [latex]\scriptsize k[/latex] multiplied by a function [latex]\scriptsize f[/latex] is the same as the constant multiplied by the derivative.
If [latex]\scriptsize f(x)=k{{x}^{n}}[/latex] then [latex]\scriptsize f'(x)=n\cdot k{{x}^{{n-1}}}[/latex] where [latex]\scriptsize n\in \mathbb{R},\text{ n}\ne \text{0}[/latex]
The derivative of a sum is equal to the sum of the derivatives.
[latex]\scriptsize \displaystyle \frac{d}{{dx}}[f(x)+g(x)]=\displaystyle \frac{d}{{dx}}f(x)+\displaystyle \frac{d}{{dx}}g(x)[/latex]
The derivative of a difference is equal to the difference of the derivatives.
[latex]\scriptsize \displaystyle \frac{d}{{dx}}[f(x)-g(x)]=\displaystyle \frac{d}{{dx}}f(x)-\displaystyle \frac{d}{{dx}}g(x)[/latex]
Applying the power rule
Let’s find the derivative of [latex]\scriptsize f(x)=2{{x}^{5}}[/latex] by applying the power rule.
We see that [latex]\scriptsize n=5[/latex] and [latex]\scriptsize k=2[/latex] by the power rule and the constant multiplied by a function rule [latex]\scriptsize f'(x)=n\cdot k{{x}^{{n-1}}}[/latex] so [latex]\scriptsize f'(x)=2\times 5{{x}^{{5-1}}}[/latex]. We multiply the coefficient [latex]\scriptsize 5[/latex] of [latex]\scriptsize f(x)[/latex] by the power of [latex]\scriptsize 2[/latex] and subtract [latex]\scriptsize 1[/latex] from the exponent to get [latex]\scriptsize f'(x)=10{{x}^{4}}[/latex].
Example 3.1
Use the rules of differentiation to find the derivative of:
- [latex]\scriptsize y=3{{x}^{{-2}}}+2[/latex]
- [latex]\scriptsize f(x)={{x}^{4}}+2{{x}^{3}}-x[/latex]
Solutions
- Remember from unit 2 there are different ways to write the derivative: [latex]\scriptsize {f}'(x)={y}'=\displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{d}{{dx}}[f(x)]=Df(x)={{D}_{x}}y[/latex]
[latex]\scriptsize \begin{align*}y&=3{{x}^{{-2}}}+2\\\therefore {y}'&=3(-2){{x}^{{-2-1}}}+0\\&=-6{{x}^{{-3}}}\end{align*}[/latex]
You will often be required to write final answers with positive exponents.
So [latex]\scriptsize {y}'=\displaystyle \frac{{-6}}{{{{x}^{3}}}}[/latex] - Here we apply the sum and difference rules.
[latex]\scriptsize \begin{align*}f(x)&={{x}^{4}}+2{{x}^{3}}-x\\{f}'(x)&=4{{x}^{{4-1}}}+2(3){{x}^{{3-1}}}-{{x}^{{1-1}}}\\&=4{{x}^{3}}+6{{x}^{2}}-1\end{align*}[/latex]
Exercise 3.1
Use the rules of differentiation to find the derivative of:
- [latex]\scriptsize y=2{{x}^{3}}[/latex]
- [latex]\scriptsize p=\displaystyle \frac{1}{2}{{q}^{4}}[/latex]
- [latex]\scriptsize g(x)=25[/latex]
- [latex]\scriptsize f(x)=2{{x}^{5}}+8[/latex]
- [latex]\scriptsize h(x)=2{{x}^{3}}-6{{x}^{2}}+3x[/latex]
The full solutions are at the end of the unit.
When to differentiate using first principles:
- If the question says ‘use first principles.’
- If we are required to differentiate using the definition of a derivative, then we use first principles.
When to use the rules for differentiation:
- If the question does not specify how to find the derivative, then we use the rules for differentiation.
Take note!
A good knowledge of exponents is essential when deriving.
Before deriving, make sure:
- There are no variables in the denominator. Any variables in the denominator should be moved to the numerator and the sign of the exponents changed.
- There are no surds. Surds should be written in the equivalent power form. For example, write [latex]\scriptsize \sqrt[3]{{{{x}^{4}}}}[/latex] as [latex]\scriptsize {{x}^{{\frac{4}{3}}}}[/latex].
- There are only separate terms (separated by plus or minus signs) in the simplified function.
- All brackets have been removed by expanding or simplifying.
Summary
In this unit you have learnt the following:
- How to apply the constant, sum and difference rules for differentiation.
- When to use the rules to differentiate.
- When to use first principles to differentiate.
Unit 3: Assessment
Suggested time to complete: 20 minutes
- Differentiate the following:
- [latex]\scriptsize y=12{{x}^{{-2}}}[/latex]
- [latex]\scriptsize y=10(7-3)[/latex]
- [latex]\scriptsize g(x)=3{{x}^{{\frac{2}{3}}}}-4x+10[/latex]
- [latex]\scriptsize g(x)=x(x+3)+5x[/latex]
- [latex]\scriptsize f(x)=\displaystyle \frac{{{{x}^{2}}-5x+6}}{{x-3}}[/latex]
- [latex]\scriptsize y=\sqrt{{{{x}^{3}}}}+\displaystyle \frac{1}{{3{{x}^{3}}}}[/latex]
- Find [latex]\scriptsize \displaystyle {{D}_{x}}\left[ {{{x}^{{\frac{2}{3}}}}-\displaystyle \frac{3}{{{{x}^{{\frac{1}{2}}}}}}} \right][/latex]
The full solutions are at the end of the unit.
Unit 3: Solutions
Exercise 3.1
- .
[latex]\scriptsize {y}'=6{{x}^{2}}[/latex] - .
[latex]\scriptsize \begin{align*}{p}'&=\displaystyle \frac{1}{2}\times 4{{q}^{{4-1}}}\\&=2{{q}^{3}}\end{align*}[/latex] - .
[latex]\scriptsize {g}'(x)=0[/latex] - .
[latex]\scriptsize \displaystyle \begin{align*}f'(x)&=\displaystyle \frac{d}{{dx}}(2{{x}^{5}}+8)\\ &=\displaystyle \frac{d}{{dx}}(2{{x}^{5}})+\displaystyle \frac{d}{{dx}}(8)&&\text{ Apply the sum rule}\\&=2\displaystyle \frac{d}{{dx}}({{x}^{5}})+\displaystyle \frac{d}{{dx}}(8)&&\text{ Apply the constant multiple rule}\\&=2(5{{x}^{4}})+0&&\text{ Apply the power rule and the constant rule}\\&=10{{x}^{4}}\end{align*}[/latex] - .
[latex]\scriptsize \begin{align*}{h}'(x)&=\displaystyle \frac{d}{{dx}}(2{{x}^{3}})-\displaystyle \frac{d}{{dx}}(6{{x}^{2}})+\displaystyle \frac{d}{{dx}}(3x)\\&=6{{x}^{2}}-12x+3\end{align*}[/latex]
Unit 3: Assessment
- .
- .
[latex]\scriptsize \begin{align*}{y}'&=-24{{x}^{{-3}}}\\&=\displaystyle \frac{{-24}}{{{{x}^{3}}}}\end{align*}[/latex] - .
[latex]\scriptsize {y}'=0[/latex] - .
[latex]\scriptsize \begin{align*}g(x)=3{{x}^{{\frac{2}{3}}}}-4x+10\\{g}'(x)=3\times \displaystyle \frac{2}{3}{{x}^{{\frac{2}{3}-1}}}-4\\=\displaystyle \frac{2}{{{{x}^{{\frac{1}{3}}}}}}-4\end{align*}[/latex] - .
[latex]\scriptsize \begin{align*}g(x)&=x(x+3)+5x\\&={{x}^{2}}+3x+5x&&\text{ Expand brackets and simplify }\\&={{x}^{2}}+8x\end{align*}[/latex]
[latex]\scriptsize \therefore {g}'(x)=2x+8[/latex] - .
[latex]\scriptsize \begin{align*}f(x)&=\displaystyle \frac{{{{x}^{2}}-5x+6}}{{x-3}}&&\text{ Factorise and cancel}\\ &=\displaystyle \frac{{(x-3)(x-2)}}{{(x-3)}}\\&=x-2\end{align*}[/latex]
.
[latex]\scriptsize \therefore {f}'(x)=1[/latex] - .
[latex]\scriptsize \begin{align*}y&=\sqrt{{{{x}^{3}}}}+\displaystyle \frac{1}{{3{{x}^{3}}}}\\ &={{x}^{{\frac{3}{2}}}}+\displaystyle \frac{1}{3}{{x}^{{-3}}}&&\text{ Get rid off roots and variables in denominator}\end{align*}[/latex]
[latex]\scriptsize \begin{align*}\text{ }{y}'&=\displaystyle \frac{3}{2}{{x}^{{\frac{1}{2}}}}-{{x}^{{-4}}}\\ &=\displaystyle \frac{3}{2}{{x}^{{\frac{1}{2}}}}-\displaystyle \frac{{1\text{ }}}{{{{x}^{4}}}}\end{align*}[/latex]
- .
- .
[latex]\begin{align*}\scriptsize \displaystyle {{D}_{x}}\left[ {{{x}^{{\frac{2}{3}}}}-\displaystyle \frac{3}{{{{x}^{{\frac{1}{2}}}}}}} \right]={{D}_{x}}\left[ {{{x}^{{\frac{2}{3}}}}-3{{x}^{{\frac{{-1}}{2}}}}} \right]&&\text{ Simplify inside brackets before deriving}\end{align*}[/latex] - .
[latex]\scriptsize \displaystyle \begin{align*}{{D}_{x}}\left[ {{{x}^{{\frac{2}{3}}}}-3{{x}^{{\frac{{-1}}{2}}}}} \right]&=\displaystyle \frac{2}{3}{{x}^{{\frac{{-1}}{3}}}}+\displaystyle \frac{3}{2}{{x}^{{\frac{{-3}}{2}}}}\\&=\displaystyle \frac{2}{{3{{x}^{{\frac{1}{3}}}}}}+\displaystyle \frac{3}{{2{{x}^{{\frac{3}{2}}}}}}\end{align*}[/latex]
