Unit 2: Work with rules for differentiation

Natashia Bearam-Edmunds

Functions and algebra: Investigate and use instantaneous rate of change of a variable when interpreting models both in mathematical and real life situations

Unit 2: Work with rules for differentiation

Natashia Bearam-Edmunds

By the end of this unit you will be able to:

Use various forms of notation to represent the derivative.
Apply the power rule.
Find the derivative of a constant.
Find the derivative of a constant multiplied by a function.
Find the derivative of a sum/difference.
Find the derivative of a product.
Find the derivative of a quotient.
Apply the chain rule.

What you should know

Before you start this unit, make sure you can:

Find the derivative by applying the constant and sum and difference rules taught in level 3 subject outcome 2.5 unit 3.

Try the following questions to see if you are ready for this unit.
Use the rules of differentiation to find the derivative of the following:

$\scriptsize g(x)=3{{x}^{{\displaystyle \frac{2}{3}}}}-4x+10$
$\scriptsize f(x)=\displaystyle \frac{{{{x}^{2}}-5x+6}}{{x-3}}$
$\scriptsize y=\sqrt{{{{x}^{3}}}}+\displaystyle \frac{1}{{3{{x}^{3}}}}$

Solutions

$\scriptsize \begin{align*}g(x)&=3{{x}^{{\displaystyle \frac{2}{3}}}}-4x+10\\{g}'(x)&=3\times \displaystyle \frac{2}{3}{{x}^{{\displaystyle \frac{2}{3}-1}}}-4\\&=\displaystyle \frac{2}{{{{x}^{{\displaystyle \frac{1}{3}}}}}}-4\end{align*}$
$\scriptsize \begin{align*}f(x)&=\displaystyle \frac{{{{x}^{2}}-5x+6}}{{x-3}}&&\text{ Factorise and cancel}\\ &= \displaystyle \frac{{(x-3)(x-2)}}{{(x-3)}}\\&=x-2\\f'(x)&=1{{x}^{{1-1}}}=1\end{align*}$
$\scriptsize \begin{align*}y&=\sqrt{{x}^{3}}+\displaystyle \frac{1}{3{{x}^{3}}}\\&= {x}^{\displaystyle \frac{3}{2}}+\displaystyle \frac{1}{3}{{x}^{{-3}}} && \text{Get rid of roots and variables in denominator}\end{align*}$
$\scriptsize \begin{align*}\text{ }{y}' &= \displaystyle \frac{3}{2}{{x}^{{\displaystyle \frac{1}{2}}}}-{{x}^{{-4}}}\\ &= \displaystyle \frac{3}{2}{{x}^{{\displaystyle \frac{1}{2}}}}-\displaystyle \frac{{1\text{ }}}{{{{x}^{4}}}}\end{align*}$

Introduction

Finding derivatives of functions by using first principles can be tedious. So, we use rules for differentiation to make the process of differentiating simpler.

As we saw in level 3 there are different notations used for derivatives. If we use the common notation $\scriptsize \displaystyle y=f\left( x \right)$ where the dependent variable is $\scriptsize y$ and the independent variable is $\scriptsize x$ , then some alternative notations for the derivative are:

$\scriptsize {f}'(x)={y}'=\displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{d}{{dx}}[f(x)]=Df(x)={{D}_{x}}y$

The symbols $\scriptsize D$ and $\scriptsize \displaystyle \frac{d}{{dx}}$ are called differential operators because they tell us that we are differentiating whenever see them.

Here is a reminder of the rules of differentiation that you should know by now.

The power rule:

If $\scriptsize f(x)={{x}^{n}}$ then $\scriptsize f'(x)=n{{x}^{{n-1}}}$ where $\scriptsize n\in \mathbb{R},\text{ n}\ne \text{0}$ .

The sum/difference rule:

$\scriptsize \displaystyle \frac{d}{{dx}}[f(x)\pm g(x)]=\displaystyle \frac{d}{{dx}}f(x)\pm \displaystyle \frac{d}{{dx}}g(x)$

In this unit we build on the differentiation rules you learnt in level 3 to include the product, quotient and chain rules. But, before we look at those there are some derivative functions that need a special mention.

The exponential and natural logarithm functions

The constant $\scriptsize e$ is a special type of exponent, known as Euler’s number, which has many real-life applications. $\scriptsize f(x)={{e}^{x}}$ is called the natural exponential function, and it is its own derivative.

$\scriptsize \displaystyle \frac{d}{{dx}}{{e}^{x}}={{e}^{x}}$

You can learn more about the uses of the natural exponent, $\scriptsize e$ , by watching the video “ $\scriptsize e$ and Compound Interest”.

$\scriptsize e$ and Compound Interest (Duration: 11.38)

The inverse of an exponential function is a logarithmic function. The inverse of the natural exponential function is the natural logarithmic function $\scriptsize \ln x$ . The base of the natural logarithmic function is $\scriptsize e$ , so $\scriptsize \ln x$ is actually $\scriptsize {{\ln }_{e}}x$ . The derivative of the natural logarithm function is $\scriptsize \displaystyle \frac{1}{x}$ .

$\scriptsize \displaystyle \frac{d}{{dx}}\ln x=\displaystyle \frac{1}{x}\text{ where }x \gt 0,x\ne 0$

Derivatives of trigonometric functions

The trigonometric functions have the following derivatives:

$\scriptsize \displaystyle \begin{align*}\displaystyle \frac{d}{{dx}}\sin x&=\cos x\\\\\displaystyle \frac{d}{{dx}}\cos x&=-\sin x\\\\\displaystyle \frac{d}{{dx}}tanx&={{\sec }^{2}}x\end{align*}$

Find the derivative of:

$\scriptsize {{e}^{x}}$
$\scriptsize y=\ln x$
$\scriptsize 2\sin x+2\cos x$
$\scriptsize f(x)=\tan x$

Solutions

$\scriptsize \displaystyle \frac{d}{{dx}}{{e}^{x}}={{e}^{x}}$
$\scriptsize {y}'=\displaystyle \frac{1}{x}$
$\scriptsize \begin{align*}\displaystyle \frac{d}{{dx}}(2\sin x+2\cos x)&=\displaystyle \frac{d}{{dx}}2\sin x+\displaystyle \frac{d}{{dx}}2\cos x &&\text{ apply the sum rule for differentiation}\\&=2\cos x-2\sin x\end{align*}$
$\scriptsize {f}'(x)={{\sec }^{2}}x$

The product rule

Now, we are ready for some of the more advanced rules of differentiation. Using the product rule we can find the derivative of the product of two functions.

$\scriptsize \displaystyle \frac{d}{{dx}}\left[ {f(x)\cdot g(x)} \right]={f}'(x)\cdot g(x)+{g}'(x)\cdot f(x)$

OR

If $\scriptsize u$ and $\scriptsize v$ are functions of $\scriptsize x$ , and $\scriptsize y=u\cdot v$ , then:

$\scriptsize \displaystyle \frac{{dy}}{{dx}}=v\displaystyle \frac{{du}}{{dx}}+u\displaystyle \frac{{dv}}{{dx}}$ .

Differentiate with respect to $\scriptsize x$ :

$\scriptsize h(x)=x\cdot \cos x$
$\scriptsize y={{x}^{2}}\cdot {{e}^{x}}$
$\scriptsize y=\ln x\cdot {{e}^{x}}$

Solutions

$\scriptsize h(x)=x\cdot \cos x$
$\scriptsize \therefore f(x)=x\text{ and }g(x)=\cos x$
$\scriptsize \displaystyle \begin{align*}\displaystyle \frac{d}{{dx}}[h(x)]&=\displaystyle \frac{d}{{dx}}\left[ {f(x)\cdot g(x)} \right]\\&={f}'(x)\cdot g(x)+{g}'(x)\cdot f(x)&&\text{ by the product rule}\\&=(\displaystyle \frac{d}{{dx}}x)\cos x+(\displaystyle \frac{d}{{dx}}\cos x)x\\&=\cos x+(-\sin x)x\\&=\cos x-x\sin x\end{align*}$
$\scriptsize y={{x}^{2}}\cdot {{e}^{x}}$
$\scriptsize u={{x}^{2}}\text{ and v}={{e}^{x}}$
$\scriptsize \begin{align*}\displaystyle \frac{{dy}}{{dx}}&=\left( {\displaystyle \frac{d}{{dx}}{{x}^{2}}} \right){{e}^{x}}+\left( {\displaystyle \frac{d}{{dx}}{{e}^{x}}} \right){{x}^{2}}\\&=2x\cdot {{e}^{x}}+{{e}^{x}}\cdot {{x}^{2}}\end{align*}$
$\scriptsize y=\ln x\cdot {{e}^{x}}$
$\scriptsize u=\ln x\text{ and }v={{e}^{x}}$
$\scriptsize \begin{align*}{y}' &=\left( {\displaystyle \frac{d}{{dx}}\ln x} \right){{e}^{x}}+\left( {\displaystyle \frac{d}{{dx}}{{e}^{x}}} \right)\ln x\\&=\displaystyle \frac{1}{x}\cdot {{e}^{x}}+{{e}^{x}}\cdot \ln x\end{align*}$

Use the product rule to differentiate with respect to $\scriptsize x$ :

$\scriptsize y=x\cdot \sqrt{x}$
$\scriptsize y={{e}^{x}}\sin x$
$\scriptsize k(x)=\ln x\cdot \tan x$
$\scriptsize h(x)=\displaystyle \frac{{{{e}^{x}}}}{x}$

The full solutions are at the end of the unit.

The quotient rule

Now that you understand the product rule, we will move onto differentiating quotients of functions. Remember that the answer to division of quantities is called a quotient. So the quotient rule is what we use to find the derivative of functions that are divided.

If $\scriptsize h(x)=\displaystyle \frac{{f(x)}}{{g(x)}}$ then $\scriptsize {h}'(x)=\displaystyle \frac{{{f}'(x)g(x)-{g}'(x).f(x)}}{{{{{\left( {g(x)} \right)}}^{2}}}}$ .

Use the quotient rule to differentiate:

$\scriptsize h(x)=\displaystyle \frac{{2{{x}^{2}}}}{{(x-3)}}$
$\scriptsize y=\displaystyle \frac{{3x-1}}{{2x+3}}$

Solutions

$\scriptsize f(x)=2{{x}^{2}}\text{ and }g(x)=x-3$
$\scriptsize {h}'(x)=\displaystyle \frac{{\displaystyle \frac{d}{{dx}}(2{{x}^{2}})\cdot (x-3)-2{{x}^{2}}\cdot \displaystyle \frac{d}{{dx}}(x-3)}}{{{{{(x-3)}}^{2}}}}$
Since the terms in the numerator are separated by a negative sign, you must be careful and use brackets to avoid mistakes.
$\scriptsize \begin{align*}{h}'(x)&=\displaystyle \frac{{4x(x-3)-2{{x}^{2}}(1)}}{{{{{(x-3)}}^{2}}}}\\&=\displaystyle \frac{{4{{x}^{2}}-12x-2{{x}^{2}}}}{{{{{(x-3)}}^{2}}}}\\&=\displaystyle \frac{{2{{x}^{2}}-12x}}{{{{{(x-3)}}^{2}}}}\\&=\displaystyle \frac{{2x(x-6)}}{{{{{(x-3)}}^{2}}}}\end{align*}$
$\scriptsize f(x)=3x-1\text{ and }g(x)=2x+3$
$\scriptsize \begin{align*}\displaystyle \frac{{dy}}{{dx}}&=\displaystyle \frac{{(2x+3)\displaystyle \frac{d}{{dx}}(3x-1)-(3x-1)\displaystyle \frac{d}{{dx}}(2x+3)}}{{{{{(2x+3)}}^{2}}}}\\&=\displaystyle \frac{{(2x+3)(3)-(3x-1)(2)}}{{{{{(2x+3)}}^{2}}}}\\&=\displaystyle \frac{{6x+9-6x+2}}{{{{{(2x+3)}}^{2}}}}\\&=\displaystyle \frac{{11}}{{{{{(2x+3)}}^{2}}}}\end{align*}$

Use the quotient rule to differentiate:

$\scriptsize h(x)=\displaystyle \frac{{{{e}^{x}}}}{{(x-1)}}$
$\scriptsize y=\displaystyle \frac{{\sin x}}{{\tan x}}$
$\scriptsize g(x)=\displaystyle \frac{x}{{{{e}^{x}}}}$

The full solutions are at the end of the unit.

The chain rule

So far we have differentiated functions using the following rules for differentiation: sum and difference rule, constant rule, power rule, product rule and quotient rule. However, these techniques do not allow us to differentiate compositions of functions, such as $\scriptsize y=\sin ({{x}^{2}})$ . To differentiate composite functions we must use the chain rule. But, what are composite functions?

You can watch this animation of the chain rule for an ‘intuitive notion of the chain rule’.

$\scriptsize y=f(g(x))$ is a composite function or a function of a function. Consider the function $\scriptsize \sin {{x}^{2}}$ . You should be able to see that this expression is different from the normal $\scriptsize \sin x$ function since $\scriptsize x$ is squared.

This is a function of function with $\scriptsize f(x)=\sin x$ and $\scriptsize g(x)={{x}^{2}}$ .

So $\scriptsize f(g(x))=f({{x}^{2}})$ .

Substitute $\scriptsize x$ with $\scriptsize {{x}^{2}}$ and you get:

$\scriptsize f(g(x))=f({{x}^{2}})=\sin {{x}^{2}}$ .

We can think of the derivative of this function as the rate of change of $\scriptsize \sin ({{x}^{2}})$ relative to the change in $\scriptsize x$ . We want to know how $\scriptsize \sin ({{x}^{2}})$ changes as $\scriptsize x$ changes. Any change in $\scriptsize x$ leads a chain reaction; as $\scriptsize x$ changes, $\scriptsize {{x}^{2}}$ changes , which leads $\scriptsize \sin ({{x}^{2}})$ to change too.

To differentiate a composite function $\scriptsize y=f(g(x))$ :

Substitute $\scriptsize u$ for $\scriptsize g(x)$ to get $\scriptsize y=f(u)$ . So $\scriptsize y$ is a function of $\scriptsize u$ and $\scriptsize u$ is a function of $\scriptsize x$ .
Apply the chain rule.

Chain rule:

If $\scriptsize y$ is a function of $\scriptsize u$ and $\scriptsize u$ is a function of $\scriptsize x$ , then $\scriptsize \displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{{dy}}{{du}}\cdot \displaystyle \frac{{du}}{{dx}}$ .

When applying the chain rule to the composition of functions, we work our way from the outside function in.

Differentiate $\scriptsize y=\sin ({{x}^{2}})$ .

Solution

Let $\scriptsize u={{x}^{2}}$ then $\scriptsize y=\sin (u)$ .

$\scriptsize \displaystyle \frac{{du}}{{dx}}=2x$ and $\scriptsize \displaystyle \frac{{dy}}{{du}}=\cos u$

From the chain rule $\scriptsize \displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{{dy}}{{du}}\cdot \displaystyle \frac{{du}}{{dx}}$ .

$\scriptsize \begin{align*}\displaystyle \frac{{dy}}{{dx}}&=\cos u\cdot (2x)\\&=2x\cos {{x}^{2}}\end{align*}$

In the last step we substituted $\scriptsize u={{x}^{2}}$ .

We can expand on the basic trigonometric derivative rules to include composite functions as shown in the box below.

$\scriptsize \begin{align*}\displaystyle \frac{d}{{dx}}a\sin kx&=k\cdot a\cos kx\\\\\displaystyle \frac{d}{{dx}}a\cos kx&=-k\cdot a\sin kx\\\\\displaystyle \frac{d}{{dx}}a\text{ }\tan nx&=k\cdot a{{\sec }^{2}}kx\end{align*}$

Find the derivative of $\scriptsize y={{(2x-1)}^{5}}$ .

Solution

You may be tempted to say just multiply out the brackets to derive, but to multiply out the brackets would take a long time and it is easy to make a mistake with so many terms. So we will treat this as a function of a function. Let $\scriptsize u=2x-1$ then $\scriptsize y={{u}^{5}}$ .

Then $\scriptsize \displaystyle \frac{{du}}{{dx}}=2$ and $\scriptsize \displaystyle \frac{{dy}}{{du}}=5{{x}^{4}}$ .

From the chain rule $\scriptsize \displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{{dy}}{{du}}\cdot \displaystyle \frac{{du}}{{dx}}$ .

So we get $\scriptsize \begin{align*}\displaystyle \frac{{dy}}{{dx}}&=5{{x}^{4}}\cdot (2)\\&=10{{x}^{4}}\end{align*}$ .

Here is a summary of the rules of differentiation for simple functions and composite functions.

Simple functions

Composite functions

$\scriptsize \displaystyle \frac{d}{{dx}}k=0,\text{ where k is a constant}\text{.}$

$\scriptsize \displaystyle \frac{d}{{dx}}k{{x}^{n}}=n\cdot k{{x}^{{n-1}}},\text{where k is a constant}.$

$\scriptsize \displaystyle \begin{align*}\displaystyle \frac{d}{{dx}}k{{[f(x)]}^{n}}&=k\displaystyle \frac{d}{{dx}}{{[f(x)]}^{n}},&&\text{ where k is a constant}\text{.}\\&=k\cdot n{{[f(x)]}^{{n-1}}}\cdot {f}'(x)\end{align*}$

$\scriptsize \displaystyle \frac{d}{{dx}}{{e}^{x}}={{e}^{x}}$

$\scriptsize \displaystyle \frac{d}{{dx}}{{e}^{{f(x)}}}={f}'(x)\cdot {{e}^{{f(x)}}}$

$\scriptsize \displaystyle \frac{d}{{dx}}\ln x=\displaystyle \frac{1}{x}$

$\scriptsize \displaystyle \frac{d}{{dx}}\ln [f(x)]=\displaystyle \frac{1}{{f(x)}}\cdot {f}'(x)$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\sin x=\cos x$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\sin f(x)={f}'(x)\cdot \cos f(x)$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\cos x=-\sin x$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\cos f(x)=-\sin f(x)\cdot {f}'(x)$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\tan x={{\sec }^{2}}x$

$\scriptsize \displaystyle \displaystyle \frac{d}{{dx}}\tan f(x)={f}'(x)\cdot {{\sec }^{2}}f(x)$

Differentiate the following functions:

$\scriptsize f(x)=\sqrt{{{{x}^{2}}+1}}$
$\scriptsize g(x)=3\tan 2x$
$\scriptsize y={{e}^{{{{{\sin }}^{2}}(2x)}}}$
$\scriptsize s=\cos {{e}^{{3t}}}$
$\scriptsize h(x)=\ln (\sin x)$

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

How to apply the product rule for differentiation.
How to apply the quotient rule for differentiation.
How to define composite functions.
How to apply the chain rule.

License

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National Curriculum (Vocational) Mathematics Level 4 by Natashia Bearam-Edmunds is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Unit 2: Work with rules for differentiation

Unit outcomes

What you should know

Introduction

The exponential and natural logarithm functions

Note

Derivatives of trigonometric functions

Example 2.1

The product rule

Example 2.2

Exercise 2.1

The quotient rule

Example 2.3

Exercise 2.2

The chain rule

Note

Example 2.4

Example 2.5

Exercise 2.3

Summary

Suggested time to complete: 35 minutes

Exercise 2.1

Exercise 2.2

Exercise 2.3

Unit 2: Assessment

License

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Unit outcomes

What you should know

Introduction

The exponential and natural logarithm functions

Note

Derivatives of trigonometric functions

Example 2.1

The product rule

Example 2.2

Exercise 2.1

The quotient rule

Example 2.3

Exercise 2.2

The chain rule

Note

Example 2.4

Example 2.5

Exercise 2.3

Summary

Unit 2: Assessment

Suggested time to complete: 35 minutes

Unit 2: Solutions

Exercise 2.1

Exercise 2.2

Exercise 2.3

Unit 2: Assessment

License

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