Differential calculus

Contents

7.2 Rules for finding derivatives

7.3 Practical applications of derivatives

7.4 Mathematical applications of derivatives Chapter summary Chapter review

Syllabus subject matter

■ Understanding of a limit in simple situations

■ Definition of the derivative of a function at a point

■ Derivative of simple algebraic functions from first principles

■ Evaluation of the derivative of a function at a point

■ Interpretation of instantaneous rate of change at a point as the gradient of a tangent and as the derivative at that point

■ Rules for differentiation including:

for rational values of n; ; ;

(product rule); (chain rule)

■ Practical applications of instantaneous rates of change

Quantitative concepts and skills ■ Basic algebraic manipulations

■ Gradient of a straight line

■ Equation of a straight line

Syllabus Guide

Chapter 7

d dp

−−−(pn) d

dr

−−− [kf r( )] d ds

−−−[f s( )+g s( )] d

dt

−− [f t( )g t( )] d dx

−−− [f g( ( )x )]

Differential calculus

Differential calculus

7.1

Finding derivatives as limits

You have seen previously that the slope of a tangent can be calculated by find the slopes of shorter and shorter chords that become more and more like the tangent. This is called a limit.

In more general terms, we describe the limit of a function as a value that it becomes close to, particularly as the value of x gets closer to some particular value. We write this as x→c. For a limit to exist, it must make no difference to the value of the function howx gets closer to c, but we do not need to make x equal to c, just as close as we like.

In Example 1, values of x on either side of 4 were used to guess the value of the function as x gets close to 4.

In this process we say that we use values of x as xapproaches 4 (x→ 4) from above (on the right of 4 on the number line) and below (on the left of 4 on the number line).

Calculus was invented independently by two different people: Isaac Newton (1642–1727) and Gottfried Leibniz (1646–1716). Although Newton was using calculus as early as 1665, Leibniz was the first to publish an account of calculus. Leibniz’s notation was superior to Newton’s and became the basis for modern calculus notation. Calculus has been called the ‘algebra of science’ because it is used so widely to link rates of change and other functions.

If it exists, the limit L of a function f(x) as x gets closer to the number c is written as = L

f x( )

x→c lim

!

Find the limit of f(x) = 3x − 5 as x gets close to 4.

Solution

We need to calculate values of the function for values of x closer and closer to 4. We do not want to know the value at x = 4, so we do not calculate this value. Construct a table of values.

The value of f(x) gets closer to 7.

Write the result. = 7

x 3 3.9 3.999 4.0001 4.1 5

f (x ) 4 6.7 6.997 7.0003 7.3 10

3 x–5

( )

x→4 lim

Example

1

3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 x

We can use factorisation or a graphics calculator to help decide whether functions with denominators that become zero have limits.

■ If the denominator can be cancelled out, the function will have a limit.

■ If the graph of the function runs off the screen of the calculator at the x value, the limit is undefined.

Find if it exists.

Solution

We can see that is not defined when x = 2.

To find the limit we need to calculate values of as x → 2, from above and below.

We can use a graphics calculator to assist with the calculations.

All calculators

Enter the function into Y₁.

Set the (or V-Window) to X_min=−2, X_max= 4, Y_min=−50 and Y_max= 50.

Draw the graph.

Casio fx-9860G AU

Select the G-Solv menu. Press and then

select Y-CAL by pressing .

Texas Instruments TI-84 and Sharp EL-9900 Select the CALC menu and then select 1:Value.

All calculators

Now enter the required X values and read off the corresponding values of Y.

In this way we can construct a table of values for as x approaches 2 from above and below.

x 1.9 1.999 1.999 99 2.0001 2.01 3

−19.1 −1999.001 −199 999 20 001 201.01 4

By looking at the values in the table, we can say that in this case there is no limit.

Write the result. is not defined.

x2_{–3 x+4}

x–2

---x→2

lim

x2–3 x+4

x–2

---x2_{–3 x+4} x–2

---WINDOW

F6 F1

x2_{–3 x+4}

x–2

---x2_–3x+4 x–2

---x2_{–3 x+4}

x–2

---x→2

lim

Example

2

Differential calculus

The formula

is used to calculate the average rate of change of a function and the gradient of a secant. Find if it exists.

Solution

We can see that is not defined when x = 5.

Write down the expression.

Factorise the quadratic. =

Cancel the common factor. = x −3 (x ≠ 5)

Since we are making x close to 5, but not actually 5, we can find the limit by finding what

x − 3 gets close to as x → 5. Write the original limit expression.

Use the result established above. =

Calculate as done previously. = 2

x2_{–8 x+15}

x–5

---x→5

lim

x2_{–8 x +15} x–5

---x2–8 x+15

x–5

---x–5

( )(x–3)

x–5

( )

---x2–8 x +15

x–5

---x→5

lim

x–3

( )

x→5 lim

Example

3

Find if it exists.

Solution

The expression is not defined when x = 6. The expression can’t be simplified. Use the graphics calculator to investigate the expression as shown in Example 2. Set the (or V-Window) with Ymin=−100 and Ymax= 100.

The graph of y = is shown on the right.

The graph goes off the scale as x → 6. Write down the expression.

Write the result. is not defined.

3 x+5

x–6

---x→6

lim

3 x+5

x–6

---WINDOW

3 x+5

x–6

---3 x+5

x–6

---x→6

lim

Example

4

f x( )₂ – f x( )₁

x₂–x₁

The diagram on the right shows two points P and Q on a curve f(x). Using this diagram, the coordinates of P are (x, f(x)), and the coordinates of Q are ((x + h), f(x + h)).

Gradient of PQ =

=

=

The gradient of a function at P is the limiting value of the gradient of the secant PQ as Q approaches P. As Q approaches P, h approaches 0.

So the gradient of the tangent at P can be written as

This expression is called the derivative of the function and has the symbol f′(x).

So f′(x) =

Note: The notation does not mean a fraction, just as the notations f (x) and f′(x) do not mean

multiplication. It is just a symbol.

Because we often make f (x) = y when graphing functions, it is also common to write the derivative in terms of x and y. Using the diagram on the right and the definition of the derivative stated previously, we can see that

f′(x)= where ∆x = h

or y′ =

=

y

x

x x + h

P

Q

f (x) f(x + h)

y = f(x)

h (run)

f(x + h) − f(x) (rise) rise

run

---f x( +h)– f x( )

x+h

( )–x

---f x( +h)– f x( )

h ---x y P Q Q′ Q′′ Q′′′ h

As Q approaches P, h approaches 0.

f x( +h)– f x( )

h

---h→0

lim

f x( +h)– f x( )

h

---h→0

lim

Derivative of a function

The derivative of the function f(x) is defined as

f′(x) = ,

provided the limit exists. The derivative can be written as

f′(x) or or f(x).

We can also define the derivative of the function f(x) at the point where x = c as

f′(c) = .

The derivative at the point where x = c can be written as f′(c) or

x = c

P

Q y = f (x)

f(x + h) − f(x)

x y

h (x + h, f(x + h))

(x, f (x))

f x( +h)– f x( )

h

---hlim→0

df dx

--- d

dx

---f x( )– f c( )

x–c

---x→c

lim df dx

---!

f(x + ∆x) − f(x) = ∆y

x y

∆x (x + ∆x, f(x + ∆x))

(x, f(x))

d dx

---f x( +∆x)– f x( )

∆x

---∆x→0 lim

∆y

∆x

---∆x→0 lim

dy dx

---The process of calculating something using the definition is known as calculating from first principles.

Alternative notation for derivative

The derivative of y = f(x) may be defined as y′ =

and written as y′ or .

∆y

∆x

---∆xlim→0

dy dx

---!

Find the derivative of 2x2− _13x + _{15 at x} = _{5 from first principles.}

Solution

Write down the definition. f′(x)=

Write the function. f (x)= 2x2 − _13x + ₁₅

Substitute (x + h) for x to find the numerator.

f(x + h) − f (x)

= 2(x + h)2− _13(x + _h) + ₁₅ − _(2x2− _13x + ₁₅₎

Expand the brackets. = 2x2+ _4xh + _2h2−_13x− _13h +₁₅−_2x2+_13x−₁₅

Simplify. = 4xh − 13h + 2h2

Factorise. = h(4x − 13 + 2h)

Replace the numerator in

the definition. f′(x)=

Cancel the common factor. =

Evaluate. = 4x − 13

Calculate derivative at x = 5. f′(5)= 7

State the result. The derivative at x = 5 is 7.

f x( +h)– f x( )

h

---hlim→0

h 4 x( –13+2h)

h

---hlim→0

4 x–13+2h

( )

hlim→0

Example

5

Find the general derivative of y = x2+ _3x − _{8 from first principles.}

Solution

Write the definition: y = f (x). f′(x) =

Write the function. f (x)= x2+ _3x − ₈

Find f(x + h). f (x + h) = (x + h)2+ _3(x + _h) − ₈

Expand the brackets. = x2+ _2xh + _h2+ _3x + _3h − ₈

Now find f(x + h) − f (x). f(x + h) − f (x)

=x2+ _2xh + _h2+_3x + _3h − ₈− _(x2+_3x−₈₎

Simplify. = 2xh + 3h + h2

Factorise. = h(2x + 3 + h)

f x( +h)– f x( )

h

---h→0 lim

The diagram shows the secant (or chord) PQ. The gradient of the secant passing through P and Q gets closer and closer to the gradient of the curve at P, as the distance between P and Q gets smaller. As the distance between P and Q

approaches zero, the gradient of the secant passing through P and Q approaches the gradient of the curve at P.

Gradient of secant PQ = mPQ =

As the distance between P and Q approaches zero, we can use limit notation to write this as

mPQ= = gradient of curve at P

However, from the work done earlier in this section, we know that

f′(x) =

So for any point on the curve y = f (x),

f′(x) = gradient of the curve at that point

Replace the numerator in the

definition. =

Cancel the common factor. =

Evaluate. = 2x + 3

Write the result: = f′(x) The derivative = 2x + 3.

h 2 x( +3+h)

h

---h→0

lim

2 x+3+h

( )

h→0 lim

dy dx

--- dy

dx

---P

Q y = f(x)

f(x + h) − f (x)

x y

h (x + h, f(x + h))

(x, f(x))

f x( +h)– f x( )

h

---Qlim→P

f x( +h)– f x( )

h

---hlim→0

f x( +h)– f x( )

h

---hlim→0

Derivative and gradient

For a function f (x), the gradient of the function at any point P(x, f(x)) is f′(x).

P

x

y (x, f(x))

Tangent

y = f (x)

!

Use the derivative to find the gradient function of f (x) = 4x2_.

Solution

Write down the formula for the derivative. f′(x)=

Write down the function. f (x)= 4x2

f x( +h)– f x( )

h

---h→0 lim

Find f(x + h) by replacing x by (x + h). f (x + h) = 4(x + h)2

Expand the brackets. = 4(x2+ _2xh + _h2₎

Expand again. = 4x2+ _8xh + _4h2

Now find f(x + h) − f(x). f (x + h) − f(x) =4x2+ _8xh + _4h2−_4x2

Simplify. = 8xh + 4h2

Factorise. = 4h(2x + h)

Substitute for f(x + h) − f(x) in the formula. f′(x) =

Cancel the common factor. =

Evaluate. = 8x

State the result. The gradient function f′(x) = 8x.

4h 2 x( +h) h

---hlim→0

4 2 x( +h)

hlim→0

Use a graphics calculator to find the gradient function of y = 7x.

Solution

Enter the function as Y1= 7X and (or DRAW) it.

Casio fx-9860G AU

Select Y2 in GRAPH mode and then press and to select CALC. Press to select d/dx.

Next press and then to select GRPH.

Press to select Y and then type:

1 0.001 and to store.

Texas Instruments TI-84

Select Y₂ and then press . Select 8:nDERIV.

Press , select Y-VARS, select 1:Function and then

select 1:Y₁. Next type:

0.001 and to store.

Sharp EL-9900

See the instructions given on the CD-ROM.

All calculators

The calculator works out the derivative numerically with an

x-interval of 0.001. the function and the derivative.

From the graph, it is clear that the gradient function is Y₂= 7.

The gradient function is = 7.

GRAPH

OPTN F2

F1 VARS F4 F1

, X,θ,T , ) EXE

MATH VARS

, X,θ,T , X,θ,T , ) ENTER

GRAPH

dy dx

---Example

8

It is possible to look at the tangents on simple curves to see how the gradient of a function changes. The spreadsheet ‘Tangent tracer’ can be loaded from the CD-ROM to investigate the way in which the tangent and its slope change.

1 Load the spreadsheet ‘Tangent tracer’ from the CD-ROM).

2 Change the function to f(x) = x2− _4x + _{1 by making a}

3= 0, a2= 1, a1= −4 and a0= 1.

3 Move the tangent along the curve.

■ Is the curve rising or falling before x = 2?

■ What happens to the slope at x = 2?

■ Is the curve rising or falling after x = 2?

4 Change the function and observe how the tangent and slope of the curve change. Note that the vertical scale changes to accommodate the curve.

Investigation

Tangents on a curve

Spreadsheet

Find which of the functions f (x) = x2+ _3x − _{5 or g(x)} = _4x − _2x2+ _{3 is steepest at x} = _5.

Solution

Find the derivative of f(x). Write the formula for the

derivative. f′(x)=

Write down the function. f (x)= x2+ _3x − ₅

Find f(x + h). f (x + h)= (x + h)2+ _3(x + _h) − ₅

Expand the brackets. = x2+ _2xh + _h2 + _3x + _3h − ₅

Now find f(x + h) − f(x). f (x + h) − f(x)= (x2+ _2xh + _h2+_3x+ _3h −₅₎ − _(x2 +_3x−₅₎

Simplify. = 2xh + h2+ _3h

Factorise. = h(2x + h + 3)

Substitute into the formula. f′(x)=

Cancel the common factor. =

Evaluate. = 2x + 3

f x( +h)– f x( )

h

---hlim→0

h 2 x( +h+3)

h

---hlim→0

2 x+h+3) (

hlim→0

Find the derivative of g(x). Write the formula for the

derivative. g′(x)=

Write down the function. g(x)= 4x − 2x2+ ₃

Find g(x + h). g(x + h)= 4(x + h) − 2(x + h)2+ ₃

Expand the brackets. = 4x + 4h − 2x2− _4xh − _2h2+ ₃

Now find g(x + h) − g(x). g(x + h) − g(x)= (4x+ 4h −2x2− _4xh − _2h2+₃₎ − _(4x−_2x2+₃₎

Simplify. =−4xh − 2h2+ _4h

Factorise. = h(−4x − 2h + 4)

Substitute into the formula. g′(x)=

Cancel the common factor. =

Evaluate. =−4x + 4

Compare the values of

f′(x)and g′(x)at x = 5.

f′(5) = 2 × 5 + 3 = 13

g′(5)=−4 × 5 + 4 = −16

Write the answer. g(x) is steeper than f(x) at x = 5.

g x( +h)–g x( )

h

---hlim→0

h(−4 x–2h+4)

h

---hlim→0

−4 x–2h+4) (

hlim→0

Show that (kx3₎ = _3kx2 _{using first principles.}

Solution

Write the formula for the derivative. f′(x)=

Write down the function. f (x)= kx3

Find f(x + h). f(x + h)= k(x + h)3

Expand the brackets. = k(x3+ _3x2_h + _3xh2+ _h3₎

Expand again. = kx3+ _3kx2_h + _3kxh2+ _kh3

Now find f(x + h) − f(x). f(x + h) − f (x)=kx3+ _3kx2_h + _3kxh2+ _kh3 −_kx3

Simplify. = 3kx2_h + _3kxh2+ _kh3

Factorise. = h(3kx2+ _3kxh + _kh2₎

Substitute for f(x + h) − f(x) in the

formula. f′(x)=

Cancel the common factor. =

Evaluate. = 3kx2

State the result in the form of the

question. (kx

3₎= _3kx2 d

dx

---f x( +h)– f x( )

h

---h→0

lim

h 3k x( 2_+3kxh+kh2)

h

---h→0

lim

3k x2_+3kxh+kh2

( )

h→0 lim

d dx

Exercise 7.1

Finding derivatives as limits

a b c

d e f

2 Use a graphics calculator to find, if possible:

a b c

d e f

3 Use algebraic methods to find, if possible:

a b c

d e f

4 Find the derivatives of the following from first principles.

a 2x + 7 at x = 4 b x − 8 at x = −2 c 3x − 5 at x = −4

d 9 at x = 3 e 12 at x = −2 f 10x at x = 6

g 9 − 5x at x = 1 h −3x + 8 at x = 7 i 7x at x = −6 1 Use a graphics calculator to draw a graph of f(x) = x2+ _3x − _{10. Adjust the}

(or V-Window) to suit.

2 Calculate f′(x) and draw the graph of f(x) and f′(x) on the same set of axes. Adjust the

to show both graphs if necessary.

3 What is the significance of the points at which f(x) cuts the x-axis?

4 What is the significance of the point at which f′(x) cuts the x-axis?

5 Clear the graphs.

6 Graph g(x) = x3− _2x2− _5x + _8.

7 Calculate g′(x) and graph it on the same set of axes as g(x).

8 What is the significance of the points at which g(x) cuts the x-axis?

9 What is the significance of the points at which g′(x) cuts the x-axis?

10 Clear the graphs.

11 Repeat steps 1 to 5 with other quadratic functions.

12 Repeat steps 6 to 10 with other cubic functions. 13 Complete the following:

The graph of the derivative of a quadratic function is a function.

The graph of the derivative of a cubic function is a function.

WINDOW

WINDOW

Investigation

Graphs of derivatives

Additional Exercise

7.1

2 x( +1) x→3

lim (4–3 x)

x→2

lim x( 2_–4)

x→5 lim

x2_–4

x–2

---x→2

lim x2–4

x+2

---x→2

lim 5

x–1

---x→3

lim

7 x( –3) xlim→4

4

x–1

---xlim→1

4

x–1

---xlim→2

x2_–25

x–5

---x→5

lim x2–16

x–5

---x→5

lim x2+x–2

x+2

---x→–2

lim

x–2

( )(x+5)

x+5

---x→–5

lim x2–36

x–6

---x→6

lim x2+2 x–3

x+3

---x→–3

lim

x2_{–2 x–8}

x–4

---x→4

lim 3 x2–8 x–3

x–3

---x→3

lim 4 x2+19 x–5

4 x–1 ---x→0.25

5 Find the derivatives of the following from the definition.

a x2 + _5x − _{7 at x} = _{3 b x}2+ _5x + _{9 at x} = _{3 c 3x}2+ _4x + _{1 at x} = −₂ d x2− _6x + _{4 at x} = _{1 e 4} − _x2+ _{2x at x =} −_{5 f 4x} − _5x2 − _6x3 _{at x} = ₈ g 4x2− _{7x at x} = _{4 h 5x}2 _{at x} = _{11 i 7x at x =} −₃

6 Find the general derivatives of the following from first principles.

a 9x + 2 b x − 7 c 3x2 _{d 9 e 11x}

f 10x2+ _3x + _{1 g 1} − _5x − _x2 _{h 3x} − _4x2+ _{2 i 7x}3 _{j x}4

7 Use first principles to find the gradient functions of the following functions.

a y =−2x3 _{b f (x)} =−_7x3 _{c f (x)} = _9x2 _{d g(x)} =−_6x2

e f (x) = x3 _{f y} = _3x2 _{g f (x)} = _12x3 _{h y} = _x

i g(x) =−4x j g(x) = 11x k y = 5x3 _{l f (x)} =−_33x2

8 Find the gradient functions of the following functions from the definition.

a 9x + 6 b x2− _7x +_{1 c 3x}2+ _8x + ₉

d 5x3 _{e 11x}4 _{f 7} − _3x2+ _9x

9 Use a graphics calculator to find the gradient functions of the following.

a 4x − 3 b 1 − 8x c x2

d 0.5x2 + _2x − _{4 e} −_x2+ _5x − _{2 f 2x}2− _3x + ₄

Modelling and problem solving

10 f(x) = 3x2− _5x + _8.

a Find the value of f′(2) from the definition.

b Find the value of f′(5) from the definition.

c Find f′(x) from the definition.

d Substitute the values 2 and 5 into the result from part c and explain the answers.

11 f(x) = 4x2_{, g(x)} = _{7x and h(x)} = _6.

a Find the general derivatives of f (x), g(x) and h(x).

b Find the general derivative of 4x2+ _7x + _6. c Work out f′(x) + g′(x) + h′(x).

d Compare the results of part b and part c and comment on the result.

12 Use the definition to show that (cx2₎ = _2cx.

13 Show that (cp) = c from first principles.

14 Use first principles to show that (k)= 0 for any number k.

15 Use the definition to show that (km3₎ = _3km2_.

16 Find which of the functions f (x) = 3x2− _5x + _{2 or g(x)} = _2x2 − _x + _{9 is steepest at x} = _3.

17 Find where the gradient function f(x) = 2x2− _4x + _{1 is higher than the gradient function of} g(x) = x2 + _8x − _3.

18 Find where the gradient function f(x) = 3x2− _5x + _{2 is higher than the gradient function of} g(x) = 3x2 + _x − _4.

d dx

---d dp

---d dk

---d dm

---7.2

Rules for finding derivatives

By now, you should have started to see that the derivatives of powers of x can be worked out using a simple rule instead of first principles.

Polynomials with more than one term can be differentiated by adding or subtracting the derivatives of the terms. Rules for differentiating polynomials are summarised as follows.

Note: The derivative of a constant is just an application of the second rule above.

Consider f(x) = c, where c is a constant. Now f(x) = cx0

So f′(x) = 0 × cx0 − 1 = ₀

Part of the track of a roller-coaster can be modelled as

h=

≈ 0.002 685d3− _0.1168d2− _0.2417d + _40.878

for 0 d 35, where d is the horizontal distance in metres and h is the height of the track in metres. The acceleration a of a roller-coaster car can be described as

a= −

where m is the gradient of the track.

Investigate points of interest with regard to the movement of the roller-coaster cars. Consider:

■ where the cars are accelerating

■ where they are decelerating

■ where they are not accelerating

■ where they might obtain the greatest speed.

It will be helpful to calculate values and draw some graphs of the model.

80d3_–3480d2_{–7200d+1 217 791} 29 791

---0.2 9.81m

m2₊₁

---+

⎝ ⎠

⎛ ⎞

Investigation

Roller-coaster speeds

Extra Material Investigating polynomial derivatives

Rules for differentiating polynomials

■ The derivative of a constant is 0. ■ The derivative of kxn _{is nkx}n − 1_.

■ The derivative of a sum (or difference) is the sum (or difference) of the derivatives.

A number of rules (theorems) can be used to differentiate functions. These allow us to find derivatives of complicated functions using relatively straightforward methods in various combinations. Each can be proven from first principles. The proofs are on the CD-ROM.

For the function f(x) = 5x3− _4x2+ _7x + _{2, find:}

a the derivative f′(x) b f′(2) c the gradient of f(x) at −1.

Solution

a Write down the function. f(x)= 5x3− _4x2+ _7x + ₂

Differentiate each term. (5x3₎= _15x2

(−4x2₎= −_8x

(7x)= 7

(2)= 0

Find f′(x) by adding derivatives of terms. f′(x)= 15x2− _8x + ₇

b Write the derivative. f′(x)= 15x2− _8x + ₇

Substitute x = 2. f′(2)= 15 × 22− ₈ × ₂ + ₇

Evaluate. = 51

c The gradient of f(x) at −1 = f′(−1). Calculate f′(−1). f′(−1)= 15 × (−1)2− ₈ ×−₁ + ₇

Evaluate. = 30

d dx

---d dx

---d dx

---d dx

---Example

11

Detailed Proof

Rules for finding derivatives

■ (k) =0 (k is a constant)

■ (kxn₎ = _nkxn − 1 _{(k is a constant) (Holds for positive and negative values of n.)}

■ [kf(x)] = kf′(x) (k is a constant)

■ [ f(x) + g(x)] = f′(x) + g′(x)

■ [ f (x) − g(x)] = f′(x) − g′(x)

Chain rule

■ [ f (g(x))] = f′(g(x))g′(x)

Product rule

■ [ f (x)g(x)] = g(x)f′(x) + f(x)g′(x)

Quotient rule

■

d dx

---d dx

---d dx

---d dx

---d dx

---d dx

---d dx

---d dx

--- f x( )

g x( )

--- g x( )f′( )x – f x( )g′( )x

g x( )

[ ]2 ---=

Some people find it easier to use the rules for calculating derivatives when these are written using an alternative notation.

The differentiation of polynomials with negative indices is shown below. Alternative notation for derivative rules

If u = f (x), v = g(x) and k is a constant:

■ y = k ⇒ = 0

■ y = kxn ⇒ = _nkxn − 1

■ y = ku ⇒ = k or y′= ku′

■ y = u + v ⇒ = + or y′= u′ + v′

■ y = u − v ⇒ = − or y′= u′ − v′

Chain rule

■ y = f(u) ⇒ = × or y′(x) = y′(u) × u′(x)

Product rule

■ y = u × v ⇒ = v + u or y′= vu′+ uv′

Quotient rule

■ y = ⇒ or y′ =

dy dx ---dy dx ---dy dx --- du dx ---dy dx --- du dx --- dv dx ---dy dx --- du dx --- dv dx ---dy dx --- dy du --- du dx ---dy dx --- du dx --- dv dx ---u v --- dy dx ---vdu dx

--- udv dx

---–

v2

---= vu′–uv′

v2

---!

Calculate the following.

a (3x−5_{) b}

Solution

a Use (kxn₎ = _nkxn − 1_{. (3x}−5₎ =−₅ × _3x−5 − 1

Evaluate. =−15x−6

b Express in the form kxn_. = _(7t−4₎

Now use (kxn₎ = _nkxn − 1_. = −₄ × _7t−4 − 1

Evaluate. =−28t−5

Express in original form. =

d dx --- d dt --- 7 t4 ----⎝ ⎠ ⎛ ⎞ d dx --- d dx ---d dx --- 7 t4 ----⎝ ⎠

⎛ ⎞ d

The derivative of a root should be treated as the derivative of a fractional index.

Now let’s see how the chain rule and product rule can be applied.

Find the derivatives of the following with respect to the given variable.

a 5x b

Solution

a Change to an index. 5x = 5x ×

Simplify the powers. _{= 5}

Find the derivative 5 = 1 × 5

Simplify and express in original form. = 7 ×

b Change to an index. =

Find the derivative. =

Express in original form. =

x 3 t2

x x x

1 2

---x1 1 2

---d dx

--- x1

1 2

--- ₁

2 --- x

1 2

---1 2 --- x

t2

3 3 _t2 _t23

---d dx

--- t

2 3 --- ₂

3 ---t

1 3

---–

2

3 t3

---Example

13

Find the derivative of y = 3(6x − 5)3_.

Solution

Write down the function. y= 3(6x − 5)3

Let u = 6x − 5. y= 3u3

Calculate . y′(u)= 9u2

Calculate . u′(x)= 6

Use the chain rule. y′(x)= y′(u) × u′(x)

Substitute for y′(u) and u′(x). y′(x)= 9u2× ₆

Substitute for u and simplify. = 54(6x − 5)2

State the result. [3(6x − 5)3_] = _54(6x − ₅₎2

dy du

---du dx

---dy dx

---Example

14

Find f′(x) if f (x) = (4x3− _3x2− _4)(4x5+ _7x).

Solution

Write as an equation. y= (4x3− _3x2− _4)(4x5+ _7x)

Let u = 4x3− _3x2− _{4 and v} = _4x5+ _{7x. y}= _u × _v

Note: This result may be expanded and written as a polynomial if desired.

Find . u′ = 12x2− _6x

Find . v′ = 20x4+ ₇

Use the product rule. y′ = vu′+ uv′

Substitute for u, u′, v, v′ and f′(x) = (4x5+ _7x)(12x2− _6x) + _(4x3− _3x2− _4)(20x4+ ₇₎

state the result.

du dx

---dv dx

---Calculate the following.

a [3(4t + 7)6_{] b}

Solution

a Write as an equation. y= 3(4t + 7)6

Let u = 4t + 7. y= 3u6

Differentiate y w.r.t. u. = 18u5

Differentiate u w.r.t. t. = 4

Use the chain rule. = ×

Substitute for and . = 18u5× ₄

Substitute for u and simplify. = 72(4t + 7)5

State the result. [3(4t + 7)6_]= _72(4t + ₇₎5

b Write as an equation. y=

Express in the form y = kxn_{. y}= _6(8u − ₃₎−9

Let z = 8u − 3. y= 6z−9

Differentiate y w.r.t. z. =−54z−10

Differentiate z w.r.t. u. = 8

Use the chain rule. = ×

Substitute for and . =−54z−10× ₈

Substitute for z and simplify. =−432(8u − 3)−10

Express as a fraction and state the result. =

d dt

--- d

du

--- 6 8u–3

( )9

---dy du

---du dt

---dy dt

--- dy

du

--- du

dt

---dy du

--- du

dt

---d dt

---6 8u–3

( )9

---dy dz

---dz du

---dy du

--- dy

dz

--- dz

du

---dy dz

--- dz

du

---d du

--- 6 8u–3

( )9

--- –432 8u–3 ( )10

Some people prefer to use the product rule with an expression involving a negative power rather than use the quotient rule. The following example shows how the expression in Example 17 can be differentiated using the product rule.

If y = , find using the quotient rule.

Solution

Write the equation. y=

Let u = 3t − 4 and v = 2t + 5. y=

Find u′. Find v′. u′ = 3 v′= 2

Use the quotient rule. y′ =

Substitute for u′, v′, etc. =

Expand brackets in numerator. =

Simplify and state the result. =

3t–4 2t +5

--- dy

dt

---3t–4 2t +5

---u v

---vu′–uv′

v2

---2t+5

( )×3–(3t –4)×2 2t +5

( )2

---6t +15–6t+8 2t +5

( )2

---dy dt

--- 23 2t+5

( )2

---Example

17

If y = , find .

Solution

Write the equation. y=

Express as a product. y= (3t − 4)(2t + 5)−1

Let u = 3t − 4 and v = (2t + 5)−1_{. y}= _u × _v

Calculate u′. u′ = 3

Calculate v′ using the chain rule. v′ =−1(2t + 5)−2× ₂

=−2(2t + 5)−2

Use the product rule. y′ = vu′+ uv′

Substitute for u′, v′, etc. = (2t + 5)−1× ₃ + _(3t − ₄₎ ×−_2(2t + ₅₎−2

Express as fractions and simplify. = −

Subtract using the LCD. =

Expand brackets in numerator. =

Simplify and state the result. =

3t–4 2t +5

--- dy

dt

---3t –4 2t+5

---3 2t+5

--- 2 3t( –4) 2t₊5)2 (

---3 2t( +5)–2 3t( –4) 2t+5

( )2

---6t+15–6t+8 2t+5

( )2

---dy dt

--- 23 2t ₊5)2 (

More care needs to be taken when differentiating complex expressions. It is accepted practice to leave differentiated expressions in factorised form unless otherwise requested.

Exercise 7.2

Rules for finding derivatives

1 Use differentiation rules to find the derivatives of the following.

a 6x4− _5x3+ _2x2+ _8x − _{9 b 9x}3+ _10x4− _3x5+ ₁₈

c x3+ _x2+ _x + _{1 d 4x}9− _8x10+ ₂₀

e 15x12+ _12x15− _36x5 _{f 5} − _x7+ _7x9+ _x10

g 5x8+ _12x9− _6x10+ _x12 _{h 11} − _3x + _x2+ _8x3− _x4

i 5x4− _17x + ₂₈ + _16x3 _{j 25x}7− _8x6+ _14x5+ _2x − ₂

2 Calculate the following.

a (3x4− _6x2+ _{18) b (3u}4− _6u2+ ₁₈₎

c (5m6+ _8m5− _24m + _{8) d (9y} + _6y5− _4y3+ _2y7₎

e (3z2− _9z3+ _38z − _{5) f (5t}7+ _9t4− _16t)

g (21 − 13r − 6r2+ _9r3_{) h (}−_6a4+ _2a2− ₁₀₎

i (8p − 3p4− _6p7− _5p9_{) j (8} + _4t2− _2t3+ _11t4₎

3 Find the derivatives of the following with respect to the given variables.

a 5x−1 _{b c} −_3x−4 _{d e 7m}−10

f g h i 16d−9 _j

Calculate the derivative of f(x) = 12(3x + 4)3_(5x − ₃₎4 _{and express it in factorised form.}

Solution

Let y = f(x). y = 12(3x + 4)3_(5x − ₃₎4

Let u= 12(3x + 4)3 _and

v = (5x − 3)4_.

y = u × v

Find u′ using the chain rule. u′ = 3 × 12(3x + 4)2× ₃

= 108(3x + 4)2

Find v′ using the chain rule. v′ = 4(5x − 3)3× ₅

= 20(5x − 3)3

Now use the product rule. y′ = vu′+ uv′

Substitute for u′, v′, etc. = (5x − 3)4× _108(3x + ₄₎2+ _12(3x + ₄₎3× _20(5x − ₃₎3

Simplify. = 108(5x − 3)4_(3x + ₄₎2+ _240(3x + ₄₎3_(5x − ₃₎3

Factorise. = 12(3x + 4)2_(5x − ₃₎3_[9(5x − ₃₎ + _20(3x + _4)]

Expand brackets. = 12(3x + 4)2_(5x − ₃₎3_[45x − ₂₇ + _60x + _80]

Simplify and state the result. f′(x) = 12(3x + 4)2_(5x − ₃₎3_(105x + ₅₃₎

Example

19

d dx

--- d

du

---d dm

--- d

dy

---d dz

--- d

dt

---d dr

--- d

da

---d dp

--- d

dt

---8

x

--- 3

t2

----3 –

u7

--- 1

y20

--- 4

v15

--- –8

z9

---Additional Exercise

4 Find the derivatives of the following with respect to the given variables.

a b c d e

f g −9z h i 4d2 _j

5 Differentiate the following

a 3(5a − 6)3 _b −_2(5x + ₄₎2 _{c 7(y} − ₂₎7 _d −_3(5q + ₈₎4

e 7(7 − 5a)3 _{f 4(5g}3+ ₃₎6 _{g 6(4x}2− _6x + ₄₎2 _h −_8(m3− _2m2+ ₄₎8

i −(9 − 3v)6 _{j 4(2g}2− _g3+ ₃₎6

6 Find the derivatives of the following.

a (x + 4)−5 _{b (3p} + ₅₎−4 _{c 2(5} − _4x)−6 _{d 7(m} + ₈₎−9

e (3p2− ₄₎−2 _{f g h}

i j

7 Find the derivatives of the following.

a (5x4− _8x3+ ₇₎5 _{b (7v}3+ _2v2+ _v − ₄₎7 _c

d e f

8 Calculate the derivative of:

a (z5− _z3+ _z − ₁₎₍₉ + _3z2− _z4₎ b (3p + 4)(p2− _2p − ₄₎

c (5v3− _3v2+ _v − _9)(7v4− _2v3− _v2+ _8v + ₆₎ d (3t2− _2t + _5)(t3+ _4t2+ _t − _6)(5t4+ _t3− _7t + ₈₎ e (4b5+ _3b2− _6b + _2)(b4+ _3b2− _6b3+ ₂₁₎ f (3u − 8u2− _6u3+ _8)(7u3− _4u2+ _9u − ₆₎

9 Find the derivative of:

a −6(v + 7)5_(v − ₆₎4 _b −₃₍₆ − _5r)3_(2r + ₉₎11

c −5(8 − 6h)7₍−₄ − _3h)6 _{d (5t} + _8)(2t − ₃₎ e −(3p + 10)7_(8p + ₁₎8 _f −₍₃ − _z)(4z + ₁₅₎6

10 Find the derivatives of the following with respect to the given variables.

a b c

d e f

11 Differentiate the following.

a (3x5− _3x2+ _8)(2x2+ _5x + ₂₎ − _(2x4+ _5x3_)(x5− ₆₎

b c 2(3m + 10)7− _5(3m + ₁₀₎4+ ₂₍₁₀ + _3m)−2

d + − e − +

f 2(3w + 5)4_(3w − ₁₎5− _(w − ₆₎8₍₁ − _3w)5 _g

x 3 m _{7 x}12--- _5t1

1 3 ---7 p 2 3 ---– 1 x

--- z5 –6u

u3

4

--- 6 d 5m2

m

---1

x+5

( )3

--- 1 3 p–4

( )5

--- 1 7 ( –z)4

---1 3m–4

( )2

--- 1 6 7( –4 x)8

---3t2_{+t –5}

6h2_–3h+1 3 _4v+5 4 _y2_{–2 y+3}

5q+9 3–5q

--- 8m–7 2m–5

--- 3 x +10 2 x–5

---t+5

t–5

--- 3 2t( –5) 4 5t( +8)

--- 12–5u 5 2u( –9)

---5c7 5 9( –4c)2

---3 2 f –5

--- 5 6 f +10

--- 4 5 f –11

--- 1

t –6

( )3

--- 4 2t+7

( )2

--- 2 3t( –5) 7 2t( +9)4

---4 p5_{–5 p}4_{+2 p}3_{– p}2_{–4 p+2} 5 p–1

( )6

---7.3

Practical applications of derivatives

The value of the derivative of a function at a given point is the slope of the tangent at that point on the graph. Since the tangent is a straight line, we can find its equation using the slope and the point.

The equation of a parabola is y = 5x − 3x2+ _{8. For the tangent:} a what is the gradient at x = 2?

b what is the equation at x = 2?

Solution

a Find the derivative. f′(x) = 5 − 6x

Find its value at x = 2. f′(2) = 5 − 12 = −7

Write the answer. The gradient of the tangent at x = 2 is m = −7.

b Find y at x = 2. y = 5 × 2 − 3 × 22 _{+ 8} = ₆

Write the values of m and (x1, y1). m = −7 and (x1, y1) = (2, 6)

Write the gradient–point formula. y − y₁ = m(x − x₁)

Substitute and simplify. y − 6 = −7(x − 2)

Expand brackets. y − 6 = −7x + 14

Write in standard form. 7x + y − 20 = 0

Write the answer. The equation of the tangent is 7x + y − 20 = 0.

Example

20

Tangent AB touches curve f(x) = 12 + 4x − x2 _{at point A and cuts the x-axis at point B.} A has an x-coordinate of 3. Find the length AB and the acute angle it makes with the x-axis.

Solution

Point A and slope of tangent

Find A. f (3) = 15, so A is the point (3, 15).

Find the slope of the tangent. f′(x) = 4 − 2x, so f′(3) = −2 = m

Equation of tangent

Write values for the formula. m = −2 and (x1, y1) = (3, 15)

Find the equation. y − y₁= m(x − x₁)

y − 15= −2(x − 3)

y − 15= −2x + 6

Write in standard form. 2x + y − 21= 0

Point B

B is on the x-axis, so y = 0.

Find B.

2x + 0 − 21= 0

x= 10.5

B is the point (10.5, 0).

Consider the function shown in the diagram at right. We saw in Chapter 5 that the average rate of change of a function f(x) is given by

=

This may also be written as

=

We saw that = instantaneous rate of change

However, in this chapter we established that

= y′= f′(x)

So, for any function f (x), the instantaneous rate of change and the derivative of the function are the same thing.

Length and angle of AB

Write the formula for length. AB =

Substitute and calculate. =

≈ 16.77

Use m = tan θ to find the angle. tan−1 ₍−₂₎ ≈−_63.4°

Write the answers. AB is about 16.77 units and the angle about 63.4°.

x₂–x₁

( )2 _y

2–y1

( )2

+

10.5–3

( )2₊(_0–15)2

P

Q y = f(x)

x y

y2− y1

x2− x1

(x2, f(x2))

(x1, f(x1)) ∆f x( )

∆x

--- f x( )2 – f x( )1

x₂–x₁

---∆y

∆x

--- y2–y1

x₂–x₁

---∆y

∆x

---∆x→0 lim

∆y

∆x

---∆x→0 lim

Instantaneous rate of change

The derivative of a function is the instantaneous rate of change of the function.

!

Find the rate of change of f (x) =−2x3− _10x + _{8 at x} = _3.

Solution

Write down the function. f (x) =−2x3− _10x + ₈

Differentiate. f′(x) =−6x2− ₁₀

Find f′(3). f′(3) =−6 × 32− ₁₀

Evaluate. =−64

State the result. The rate of change of f (x) at x = 3 is −64.

The volume of liquid in an underground tank being gravity-filled from a tanker is given by

V = 1200 + 1020t − 17t2

where the volume V is in litres and the time t is in minutes.

a What is the filling rate after 5 minutes?

b What is the flow rate in the pipe after 10 minutes?

Solution

Both the filling rate and the flow rate are given by the instantaneous rate of change of the volume, so we use the derivative.

Calculate the derivative. = 1020 − 34t

a Find value of derivative at t = 5.

t ₌ 5 =

1020 − 34 × 5

Evaluate. = 850

State the result. The filling rate after 5 minutes is 850 L/min.

b Find value of derivative at t = 10.

t ₌ 10=

1020 − 34 × 10

Evaluate. = 680

State the result. The flow rate after 10 minutes is 680 L/min.

dV dt

---dV dt

---dV dt

---Example

23

The volume of a sphere is given by V = .

1 Calculate the volume of a sphere of radius 8 m.

2 Calculate the volume of a sphere of radius 8.1 m.

3 What effect does an error of 0.1 m in the radius have on the volume of a sphere (from these calculations)?

4 Calculate the effect on the volume of an error of 1 m in the radius (from 7.5 to 8.5 m).

5 Calculate the value

r ₌ 8

. What unit does this value have?

6 Compare your answers to questions 4 and 5.

r 4πr3

3

---dV dr

Exercise 7.3

Practical applications of derivatives

1 f (x) = 2x2+ _x − ₄

a Find the gradient of the tangent at x = −2.

b Find the equation of the tangent at x = −2.

c Find the acute angle the tangent at x = −2 makes with the x-axis.

d Find the gradient of the tangent at x = 0.

e Find the equation of the tangent at x = 0.

f Find the acute angle the tangent at x = 0 makes with the x-axis.

2 y = x2− _2x3+ _5x − ₁

a Find the equation of the tangent at x = −2.

b Find the gradient of the tangent at x = 1.

c Find the equation of the tangent at x = 1.

d Find the acute angle the tangent at x = 1 makes with the x-axis.

3 p = 3x2− _5x + ₂

a Find the acute angle the tangent at x = 1 makes with the x-axis.

b Find the length of the tangent at x = 5 cut off by the axes.

c Find the length of the tangent at x = −3 cut off by the axes.

d Find the acute angle the tangent at x = −3 makes with the x-axis.

4 P = t2 _{+ 4t} − _4t3 − ₄

a Find the gradient of the tangent at t = 0.

b Find the acute angle the tangent in part a makes with the t-axis.

c Find the equation of the tangent at t = −2.

d Find the acute angle the tangent in part c makes with the t-axis.

e Find a general expression for the gradient of the tangent.

f Find the point(s) where the gradient of the tangent is zero.

g What is the slope of the normal at the points found in part f?

h Find the equation of the tangent at t = 3.

i Find the gradient of the tangent at t = 1.

j Find the equation of the normal at t = 1.

Investigation continued

7 What would you expect the approximate error to be if there were a 0.4 m error in the radius (from 7.8 to 8.2 m)?

8 Repeat questions 4 to6 for r = 50 m (error of 1 m).

9 Use the derivative to calculate the approximate error in volume if a 1 m error is made in the radius and r ≈ 700 m.

10 What approximate error in the volume would occur if a mistake of 0.2 m were made in the radius at a radius of 700 m?

Additional Exercise

5 f(x) = x3 − _4x2 + _x + _{3. Find the equation of the normal at:}

a x = 2 b x = 1 c x = −2 d x = −3

6 Find the rate of change of y = 7x3+ _2x2− _12x + _{5 at:}

a x = 1 b x =−2 c x = 3 d x = 0 e x = 5

7 Find the rate of change of E = 2x2+ _4x3− _{18x at:}

a x =−2 b x = 3 c x =−1 d x = 0 e x =−4

8 s = t2− _2t3+ _5t − ₆

a Find the average rate of change from t = 2 to t = 8.

b Find the instantaneous rate of change at:

i t = 2 ii t = 8 iii t = 5

c Compare the average rate of change and the instantaneous rates of change.

Modelling and problem solving

9 q(x) =

a What is the domain of q(x)?

b Find the slope of the tangent at x = 13.

c Find the equation of the tangent at x = 13.

d Find the slope of the tangent at the point A where x = a.

e Find the slope of the line from A to the origin.

f Hence find the point A on the curve such that the tangent passes through the origin.

10 y =

a Find the slope of the tangent at x = −2.5.

b Find the slope of the tangent at x = −1.5.

c What can you say about the tangents at x = −2.5 and x = −1.5?

d Find the slope of the tangent at x = −2.2.

e Find the slope of the tangent at x = −1.8.

f What can you say about the tangents at x = −2.2 and x = −1.8?

g Find the slope of the tangent at x = −3.

h Find the slope of the tangent at x = −1.

i What can you say about the tangents at x = −3 and x = −1?

j Write a general expression for the x-coordinates of points on this curve that have parallel tangents.

11 The approximate height of a pebble thrown into a well is given by

h = 2 + 9t − 5t2

where height h is in metres and time t in seconds.

a Draw a graph showing the height from 0 s to 3 s in 0.2 s intervals.

b Calculate the speed of the pebble.

c Draw a graph of the speed of the pebble from 0 s to 3 s in 0.2 s intervals.

d Calculate the average speed from 0 s to 3 s.

e Compare the average speed to the instantaneous speed.

x–4

1

x+2

---+2

dh dt

---12 Refer back to the investigation ‘Errors and derivatives’ which precedes this exercise.

a Use derivatives to find the approximate error made in the volume of a cylindrical water tank of radius 4 m and height 3 m if the radius has an error of 0.09 m.

b What approximate error would be made in the volume of the water tank above if an error of 0.09 m were made in the height instead of the radius?

13 An iceberg approximates the shape of a rectangular prism and is about 800 m long, 400 m wide and 120 m thick. As it moves into warmer water it melts evenly by 1 m per day along each measurement.

a Find the volume of the iceberg after

t days.

b Find the rate of change of the volume after t days.

c How much ice melts on the 13th day?

d How much melts on the 15th day?

14 An ant-lion makes a hole in the sand in the shape of an inverted cone and sits in the bottom of the hole. The slope of the sides is 45º. When an ant falls into the trap, the ant-lion gets its reward but has to redig the hole.

a Write an expression for the volume of the hole in terms of its depth.

b What is the rate of change of volume at depth d?

c If the hole is 1.5 cm deep, how much sand must the ant-lion dig to make it 1.7 cm deep?

15 In economics, the marginal cost of production is the additional cost of a small increase in production. It is the rate of change of the cost function (i.e. cost per item) as production increases. The cost function for the production of ovens is given by

C(x) = 0.04x3− _10x2+ _1200x + ₂₀₀₀

where the cost C(x) of producing x ovens is given in dollars.

a Find the cost of producing 60 ovens.

b Find the average cost of producing each oven if 60 ovens are made.

c Use the derivative to find the marginal cost of producing an extra oven at 60 ovens.

d Compare the average and marginal costs at:

i 60 ovens ii 100 ovens iii 130 ovens.

7.4

Mathematical applications of

derivatives

The gradient of a curve can be used to determine how a function is changing. For an increasing function, as the x values increase, the y values also increase. For a decreasing function, the

Some points where functions are increasing, decreasing or stationary are shown on the following diagram, with the corresponding gradient functions underneath.

Decreasing

Negative

Negative Decreasing Increasing

Stationary

Stationary

Functions

y

x

y

x

Stationary Increasing

Stationary

Stationary Increasing

Decreasing Increasing

Positive

Positive

Positive

Negative Positive y

x y

x

Gradient functions

Function and gradient function

■ If a function is increasing, the gradient will be positive. The gradient function is above the x-axis.

■ If the function is decreasing, the gradient will be negative. The gradient function is below the x-axis.

■ We say that the function is stationary at points where the gradient is zero. The gradient function intersects the x-axis.

!

Sketch a possible gradient function for the function shown on the right.

Solution

The function is at first increasing, so the gradient function is above the x-axis (positive).

There is a stationary point just before the y-axis, so the gradient function has an x-intercept.

Between the top and bottom of the curve, the function is

decreasing, so the gradient function is below the x-axis (negative).

The gradient function again cuts the x-axis at the x value where the second stationary point occurs.

After the second stationary point, the function is increasing, so the gradient function is above the x-axis (positive).

x y

y

x

State whether the function k(n) = 2n2− _4n + _{4 is increasing, de}