unit 1- the derivative

8/2/2019 Unit 1- The Derivative

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UNIT 1

The Derivative

Overview

In this unit we will examine in depth the problem of finding the exact value of the

gradient of a curve. You will see the deficiency of estimating the gradient by drawing a

tangent by eye, and discover that an exact value can be found by calculation once the

function that determines the shape of the curve is known. Do not be afraid of the word

Calculus it is a beautiful topic with almost limitless applications.

Learning Objectives

After you have completed this unit, you should be able to

1. Calculate the gradient of a straight line given the coordinates of two points on theline.

2. Estimate the gradient of a curve at a pointPby drawing the tangent atP, thenfinding the gradient of the tangent.

3. Calculate exactly the gradient of the tangent to the curve y = ax2 +bx +c at anygiven point on the curve.

4. Determine whether or not a given sequence has a limiting valueL, and findLwhere it exists.


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ME 22A Unit 1 2

5. Calculate the limiting value of a stated function ofx as x tends to a given limitingvalue.

6. Determine whether a given function f(x) is continuous at some stated value ofx,and find any discontinuities that may exist.

7. Differentiate f(x) = axn (a, n are constants) from first principles.8. Use the result f '(x) = naxn-1 to differentiate functions of the form f(x) = axn and

any sums or differences of such functions.


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ME 22A Unit 1 3

Session 1.1

The Gradient of a Curve

Before we begin dealing with the meat of this course, there are certain prerequisites that

are essential, notably the definitions of the gradient of a line and the gradient of a curve.

Gradient of a Line

You have almost certainly met this concept before, either in a mathematics course or one

of the sciences or even in geography.

Fig. 1

IfAC and BC are drawn parallel to the axes, the gradient ofAB is defined as the ratio

BC/AC, i.e.

the gradient ofAB = BC/AC.

Clearly,AC = x2 - x1 andBC = y2 - y1.

Therefore the gradient ofAB =y 2y1x2 - x1


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4 ME 22A Unit 1

Example

The gradient of the line joining the points (- 1, 3) and (2, - 2) is

3 (2)

12=

3 +2

3=

5

3

Gradient of a Curve

Fig. 2

The gradient of a curve y = f(x) at a pointPis defined as the gradient of the tangent to the

curve atP. Once again, you have probably come across this idea before, whether in a

ACTIVITY 1.1

Find the gradient of the line joining the points

(i) (- 2, - 6) and (3, 4),(ii) (1, - 8) and (3, - 6).


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ME 22A Unit 1 5

mathematics course or possibly (for example) in a chemistry course when finding a rate of

reaction.

The question now is how do we find the gradient of this tangent?

Finding the Gradient of a Curve

We will approach the task of finding the gradient of a tangent to a curve by considering a

particular problem (a problem which we will only manage to solve at the very end of the

course!). The problem is:

The number of microbes (n) in an organism after a time tis such that n = 2 t , where tis

given in minutes. The graph ofn = 2 t is shown below (Fig. 3). What will be the rate of

increase ofn when t= 3?

Fig. 3


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6 ME 22A Unit 1

You may (or may not) know that the rate of increase ofn can be found by drawing the

tangent to the curve at the pointP(where t= 3) and finding its gradient. More work will

be done on rates of change later in the course.

How would you draw this tangent? If your answer is to draw it by eye (so to speak), you

may not be too satisfied with this approach. You must admit that the method is rather hit

or miss!

Some people, in order to draw the tangent as accurately as possible, take two points on the

curve which are equidistant fromP(e.g. A and B in Fig. 4 below).

Fig. 4

ACTIVITY 1.2

Draw the tangent by eye and estimate its gradient.


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ME 22A Unit 1 7

The tangent atPis then drawn parallel to the chordAB.

As a matter of interest, the value of the gradient (to 4 significant figures) is 5.545. How

close were you to this answer?

[By the way, this is not the final solution to our original problem! What we are seeking is the exact

value of the gradient.]

Can you spot any defects in this technique? One criticism is that the method assumes

symmetry of the curve about the point P(at least in the regionA toB). Another fault, of

course, is that the method still involves a construction and can therefore never achieve

the degree of accuracy obtained by using a calculation. If we need an accurate (even anexact) answer, therefore, what are we to do?

Finding an Accurate Value for the Gradient

It is of interest to note that with some curves accurate tangents can be constructed. One

interesting curve for which this is possible is the parabola

y = ax2+bx +c

Consider, for example, the curvey = 5 + 4x 2x2. A sketch of this graph is drawn in Fig. 5.

ACTIVITY 1.3

What value would you obtain for the gradient using this method?

Does it differ from your previous answer?


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8 ME 22A Unit 1

Fig. 5

What would be the gradient of this curve when x = 2 (for example)?

The pointPcorresponds to the point on the curve where x = 2 . We now draw a lineBC

parallel to the x-axis through the maximum pointB, and a line throughPparallel to they-

axis to meetBC atA. If the mid-pointMofAB is now found, the required tangent isMP

(believe it or not!).

From the above drawing (Fig. 5), the required gradient would be

-AP

AM= -

212

= 4.

One wonders whether this answer would have been obtained when sketching the tangent

by eye!

[As a matter of interest, a proof of this method will be given at the end of Unit 1].


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ME 22A Unit 1 9

Although the above result concerning the function ax2+bx + c isinteresting, how can we

find an accurate value for the gradient of a tangent with other functions? In particular

(reverting to our original problem), how can we find the exact gradient of the tangent to

the curve n = 2t, when t= 3? This is where the beauty of the Calculus comes in!

ACTIVITY 1.4

Fig. 6

0x

y

8

9

7

6

5

4

3

2

1

1 2 3 4

A sketch of the graph ofy = x22x + 5 is shown above. Calculate the exact value of

the gradient of the tangent at the point (3, 8).


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10 ME 22A Unit 1

Exercise 1.1

1. Find the gradient of the line segment joining the points

(i) (7, 5) and (- 1, 1),

(ii) (- 2, - 3) and (- 7, 1).

2. Draw the graph ofy = x3 for - 1 x 2 and estimate the gradient of the curve at thepoint where x = 1.5.

3. Find the exact value of the gradient of the curvey = x2 - 6x + 5 at the point (1, 0).

The answers to these questions will be given in the Appendix.


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ME 22A Unit 1 11

Session 1.2

Differentiating From First Principles

As with most brilliant ideas, the concept of finding the gradient of a curve by means of the

Differential Calculus is based on a very simple notion. Consider the curve y = f(x) drawn

below :

Fig. 7

If we are asked to find the gradient of this curve at the point P(x, f(x)), we first consider a

point Q with coordinates (x + h, f(x + h)) which is close to P, i.e. h is small. Note that hcould be positive or negative, i.e. Qcould be on either side ofP. The gradient of the chord

PQis obviously not equivalent to the gradient of the tangent at P, but as Qis moved closer

and closer toP, the gradient ofPQmoves closer and closer to the gradient of the tangent

atP, i.e.

Fig. 8


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12 ME 22A Unit 1

But the gradient of the chord PQ = f(x+ h) - f(x)

x + h x=

f(x + h) f(x)

h

Therefore, as Qmoves closer and closer toP(we write Q

P ),

f(x +h) f (x)

h the gradient of the tangent atP. We can therefore say that the limiting

value off(x + h) f (x)

has h 0 (i.e. Q P ) gives the gradient of the tangent atP, i.e.

the gradient of the tangent at P = limh 0

f(x + h) f (x)

h

If we write f '(x) to denote the gradient of the tangent atP(x, f(x)), where f '(x) is termed the

derivative off(x), then

f'(x) = limh0

f(x +h) f(x)

h

Numerical Example

Let us consider the problem of finding the gradient of the curve y= x2 at the pointP(2, 4).

Fig. 9


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ME 22A Unit 1 13

Suppose we take another point Qon the curve where x = 21/2. The coordinates ofQwould

then be (2.5, 6.25) and the gradient ofPQis6.254

2.52=

2.25

0.5=4.5 Let us now move Qa

little closer toP, say to Q1 (2.25, 5.0625). The gradient ofPQ1 =5.0625 4

0.25=4.25. Moving

now to Q2 (2.1, 4.41), the gradient of PQ2 = 4.41 40.1

= 4.1. Going even closer, if the

coordinates ofQ3, Q4, Q5 are (2.01, 4.0401), (2.001, 4.004001), (2.0001, 4.00040001)

respectively, the gradients ofPQ3, PQ4, PQ5 would be 4.01, 4.001, 4.0001.

The values of the gradients we obtained (as Qmoved closer toP) were 4.5, 4.25, 4.1, 4.01,

4.001, 4.0001. It is not difficult to see, therefore, that the gradient is approaching a value

of 4 and will get closer and closer to 4 as Qgets closer and closer toP. In fact, if we take Q

close enough toP, the difference between the gradient ofPQand 4 can be made smaller

than any arbitrary value we care to name.

We therefore conclude that the gradient of the curve y = x2 at the pointP(2, 4) would be

4, which is the correct answer. (We will confirm this result in a few moments).


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14 ME 22A Unit 1

ACTIVITY 1.5

Fig 10

Examine the gradient of the chordPQas QP, wherePis the point (1, 1) on the curve y = x3

(shown above). Taking the coordinates ofQas (1.25, 1.953), the gradient of

PQ=1.953 1

1.25 1= 3.812

Taking Qmoving closer toP, i.e. taking the coordinates ofQ1, Q2, Q3, Q4 as 1.1, 1.01, 1.001,

1.0001, what would be the corresponding values of the chord PQ? What do you think is the

value of the gradient atP?


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ME 22A Unit 1 15

Differentiating from First Principles

Taking once again the function f(x) = x2

, the derivative f '(x) is given by

f'(x)=

limh0

f(x+h )f(x )

h

{ }=

limh0(x+h)2 x2

h

{ }

= limh0

x2 +2hx+h 2 x2

h{ } = limh02hx+h 2

h{ }

= limh0

(2x +h)= 2x.

To find the gradient at the point (2, 4) we then substitute x = 2 , i.e. the gradient is 2 x 2 =

4 (as we obtained earlier).

Example 1Find f '(x) when f(x) = 2x3. Hence find the gradient of the tangent at the point (3, 54).

f'(x) = limh0

f( x+h)f( x)

h{ } = limh02( x+h)32x3

h

= lim

h0

2( x3 +3x2h+3xh2 +h 3 )2x3

h

= limh0

6x2h +6xh2+ 2h3

h{ } = limh0(6x2

+ 6xh+ 2h2

) =6x2

.

The gradient of the tangent when x =3 is then 6 x 32 = 54 .

Example 2Referring to the problem of finding the gradient of the tangent to the curve

y = 5 + 4x 2x2 at the point where x = 2 (see Fig. 5),

f'(x) = limh0

f(x+h) f(x)

h{ } = limh05+4(x+h)2(x+h)

2 5+4x2x

2]h

= limh0

5+4x+4h 2(x2+2hx+h 2 )54x+2x2

h

= lim

h0

4h4hx2h2

h{ }


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16 ME 22A Unit 1

= limh0

(4 4x 2h) =4 4x.

Therefore, ifx = 2, the gradient of the tangent = 4 4 x 2 = - 4 (as obtained earlier).

Example 3Find f '(x) where f(x) = 1/x. Hence find the gradient of the curvey = 1/x at the point (2, 1/2).

f'(x) = limh0

f( x+h)f( x)

h{ } = limh01

(x+ h) 1x

h

= lim

h0

x(x+h)

hx( x+h){ }

= limh0

/h1

/h1x(x+h)

= 1

x2.

Therefore, ifx = 2, the gradient = - 1/4.

It is probably becoming increasingly obvious to you, however, that calculating f'(x) from

first principles is rather tedious! You are no doubt asking Is there an easier method?

Please! You will be glad to know that the answer is Yes. Before dealing with this result,

however, (i.e. developing a formula for the differentiation of axn, where a and n are

constants), we will look a little more deeply into the concepts of limits and continuity.

ACTIVITY 1.6

Differentiate from first principles

(i) f(x) = x2 + 1, (ii) f(x) = 1/x2, (iii) f(x) = x.

Hence find

(iv) the gradient of the tangent to the curve y = x2 + 1 at the point (- 1, 2),

(v) the gradient of the tangent to the curve y = xat the point (9, 3).


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ME 22A Unit 1 17

Exercise 1.2

1.

Fig. 11

Taking x-values ofQas 1.1, 1.01, 1.001, 1.0001, and finding the corresponding

gradients of the chordPQ, show that the gradient of the tangent at P(1,1

4) to the

curve y=1

1+ x( )2

is 1

4.

2. Referring to the curve n = 2t and our problem of finding the gradient at P(where t

= 3), take points Q1, Q2, Q3 and Q4 close toPwith t-coordinates 3.1, 3.01, 3.001 and

3.0001. Find the gradients of the chords PQ1, PQ2, PQ3 andPQ4. Do the magnitudes

of these gradients get closer and closer to 5.545 as we suggested earlier in session

1.1?

3. Differentiate from first principles

(i)1

x2, (ii) x

23x +6 , (iii)

1

x.

Find the gradient of the tangent to the curve y=1

x2at the point (1, 1).


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ME 22A Unit 1 19

Session 1.3

Limits and Continuity

LIMITSLooking back once again to the sequence of values 4.5, 4.25, 4.1, 4.01, 4.001, 4.0001 that

we obtained earlier, it is not difficult to see that, if the pattern of the last four terms is

continued, the sequence would proceed in the form 4.00001, 4.000001, 4.0000001, etc.

Clearly, we could continue until we reach as close to 4 as we wish without actually

reaching 4. For example, if we require the difference to be less than 10-10

, i.e. 1/1010

, wecould continue until we reach 4.00000000001. The limiting value of the sequence is then

said to be 4.

A more formal definition of a limit is given below:

A sequence of values u1, u2, u3,, un, would tend to a limitL if the difference between

un andL can be made smaller than any arbitrarily small value of that we care to

state, i.e. | un -L | < , where 0.

ACTIVITY 1.7What is the limiting value of the sequence

(i) whose nth term is 12( )

n,

(ii)

whose nth term is 4+

1

n ,

(iii) whose nth term is 2n +3n +1

(iv) 0.1, 0.11, 0.111, 0.1111, ?


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20 ME 22A Unit 1

In most cases limxa

f(x) can be found by simply substituting x = a. For example,

limx1

2x+3( ) =5. However, if putting x = a results in obtaining lim (0/0), clearly this is

undefined. For example, limh0

(x +h)2 x 2

h

would give lim (0/0) if we simply put h = 0.

In this kind of situation the function needs to be re-written, i.e. limh0

(x+h) 2 - x2

h

=

limh0

x2 +2hx+h 2 - x 2

h

= lim

h0

2hx +h 2

h

= lim

h0(2x+h) =2x.

Examples on Limits

Evaluate the following limits

(i) limx 1

3x 2

x +1

, (ii) lim

h05x

2+2hx +h

2( ), (iii) limh0

3x +h( ) x h( )[ ],

(iv) limx 2

2x

2

3x 2x 2 4

, (v) lim

h0

1

x+h( )2 1

x2

h

The solutions would be

(i)3 2

1+1=

1

2, (ii) 5x2 (simply put h = 0), (iii) 3x2 (simply put h = 0),

(iv) limx 2

2x +1( ) x 2( )x +2( ) x 2( )

= lim

x 2

2x +1x +2

=

54

(v) limh0

x2 x +h( )

2

hx2

x +h( )2

= lim

h0

x2 x2 2hx h 2

hx2

x +h( )2

= lim

h0

2hx h2

hx2

x +h( )2


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ME 22A Unit 1 21

= limh0

2xh

x2

x +h( )2

= 2x

x4=

2

x3.

ACTIVITY 1.8Evaluate the limits

(i) limx 3

2x 1

x +2

, (ii) limh0

5x h( ) 2x +h( )3x h

, (iii) lim

x 4

3x2 +11x 4

x 2 16

(iv) limh0

1

x+h( ) 1

x

h

.

Rules Regarding Limits

The solution to (ii) above can be found by re-writing limh0

5x2+2hx +h

2( ) as

limh0

5x2( ) + lim

h02hx( ) + lim

h0h

2( ) =5x2 +0 +0 =5x 2.

Similarly, the solution to (iii) can be found by re-writing limh0

3x +h( ) x h( )[ ]

as

limh0

3x +h( ) limh0

x - h( ) =3x x( ) = 3x2.


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The above results serve in part as illustrations of the general rules

Rule 1 limxa

kf x( )[ ]= k limxa

f x( )[ ],

Rule 2 limx a f x( )g x( )[ ]=

limx a f x( )[ ]

limx ag x( )[ ],

Rule 3 limx a

f x( ) g x( )[ ] = limxa

f x( )[ ] limxa

g x( )[ ],

Rule 4 limx a

f x( )g x( )

= lim

xaf x( )[ ] lim

xag x( )[ ], if lim

xag x( )[ ] 0.

CONTINUITY

Fig. 12

A functionf(x) is continuous at the point wherex = a, if

1. f(a) exists, and2. f a +h( ) f a( ) as h 0 (no matter whether h > 0 or h < 0).


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ME 22A Unit 1 23

Discontinuities

With some functions obvious discontinuities exist, for example f x( ) =1

1+x( ). Clearly f(-1)

has no value, as division by zero is not possible. The graph ofy= 11+ x( )

is shown below

Fig. 13

The graph shows a clear discontinuity where x = - 1.

ACTIVITY 1.9

For the following functions, find the values of x at which discontinuities occur:

(i)3

x 1, (ii)

3x 1

x2 7x +6.

With some functions, however, the question of whether or not continuity exists at certain

specific points is not quite so obvious. Consider the following examples.


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24 ME 22A Unit 1

y

x-2 -1 0 1 2

3

2

1

1

0

-1

y

x

Example 1

Is the function f(x) = | x | , i.e. the modulus ofx, continuous at x = 0?

The graph ofy = | x | is drawn below, where | x | = x if x 0, but = - x if x < 0.

Fig. 14

Does this function fulfil the requirements of continuity at x = 0? First of all, the function is

defined when x = 0, i.e. | 0 | = 0. Secondly, f 0 +h( ) f 0( ) as h 0 no matter whether h 0. The function is therefore continuous at x = 0 (and at every other point for that

matter).

Example 2

Is the function fdefined below continuous at x = 0?

f(x) = + 1 when x 0, f(x) = - 1 when x < 0.

The graph ofy = f(x) is drawn below.

Fig. 15


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ME 22A Unit 1 25

The function is defined when x = 0, i.e. f(0) = + 1, so it fulfils the first condition. However,

f 0 +h( ) +1 as h 0 when h > 0, but f 0 +h( ) 1 as h 0 when h < 0. Therefore we

must conclude that the function has a discontinuity at x = 0 .

Continuity and Differentiability

From the method we used to develop the derivative of a function, i.e. taking the limiting

value of the gradient of the chordPQas QP(and hence as f x +h( ) f x( ) ) , it should be

clear that a function must be continuous at a point x = a in order to have a derivative at

x = a.

However, the converse of the above result is not necessarily true, i.e. a function can be

continuous at x = a, but have no derivative at x = a. For example, the function f(x) = | x | iscontinuous at x = 0 (as we have already seen), but it has no derivative at this point. For

one thing, the gradient of the function is + 1 on one side of the point where x = 0, and 1

on the other side.

After our brief digression on the topics of limits and continuity, we will now return to our

task of finding a formula to differentiate the function axn (where a and n are constants).


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26 ME 22A Unit 1

Exercise 1.3

1. Which of the following sequences would have a limiting value L? Find L in those caseswhere the limit exists.

(i) The sequence whose nth term is 3 12( )

n

.

(ii) The sequence whose nth term is 3(2) n.

(iii) The sequence whose nth term isn 1

2n +1

2. Evaluate the limits

(i) limx1

2

2x 2 + 9x - 5

2x2 x

, (ii) limh0

x + h[ ]3

- x3

h

,

(iii) limh0

11 + x + h( )

- 11 + x( )

h

3. At which value of x does the function 1(x 1)

have a discontinuity?

Where does the curve y=1

x 1( )intersect the y-axis?

What happens to the function when x or x - ?

Sketch the curve.


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ME 22A Unit 1 27

Session 1.4

Derivative of axn

We will now make life a little easier for you by showing that, iff(x ) = axn (a, n are

constants), then f '(x) = naxn-1, e.g. iff(x) = 3x6 , f'(x) = 18x5 .

In order to develop this proof, we need the result

xn - an = (x a)(xn-1 + axn-2 + a2xn-3 + a3xn-4 + + an-1).

If you have met algebraic long division before, the factorization can be explained as

follows:

x a xn

xn ax n1

ax n1

axn1

a2x

n2

a2x

n2

.

.

.an-1x - an

an-1x - an

0

xn1 +ax n2 +a2xn3 +K+an1

)

However, the result is easily explained once (x a)(xn-1 + ax n-2 + a2x n-3 + + a n-1) is

expanded, giving

x n + ax n-1 + a2x n-2 + a3x n-3 + + an-1 x

- ax n-1 - a2x n-2 - a3x n-3 - - an-1 x - an

= xn - an .


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28 ME 22A Unit 1

Example

x4 - a4 = (x a)(x3 + ax2 + a2x + a3).

Differentiating axn (a, n are constants) from first principles

If f(x) = axn ,

f'(x) = limh0

f x +h( ) f x( )h

= lim

h0

a x +h( )nax

n

h

= limh0

a /x+ /h1

/x

x+h[ ]

n1+x x+h[ ]

n2+ x

2x+h[ ]

n3+K+x

n1

/h1

= limh0

a x +h[ ]n1

+x x+ h[ ]n2

+ x2

x+h[ ]n3

+K+xn1

=ax

n1 n =anx

n1

Example

Differentiate the following functions using the resultd

dxax

n[ ] = naxn1.

(i) 5x10 , (ii) 16x4, (iii)3

x2, (iv) x , (v)

2

x3

,

(vi) 3x , (vii) 2 .

[Note that examples (iii) to (v) assume some knowledge of indices, i.e. xn

=

1

xn,

x1n= x

n ]


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ME 22A Unit 1 29

(i) ddx

5x10[ ] = 50x9

(ii) ddx

16x4[ ] =64x3

(iii) ddx

3

x 2

=

d

dx3x

-2[ ] = -6x -3 = 6x3

(iv) ddx

x[ ] = ddx x12

=

1

2x

- 12=

1

2 x( )

(v) ddx

-2

x3

=

d

dx-2x

-13

=

2x

3

43

=

2

3 x3( )4

(vi) ddx

3x[ ] =d

dx3x1[ ] = 3x0 =3 1= 3.

(vii) ddx

2[ ] =d

dx2x0[ ] = 0.

ACTIVITY 1.10Differentiate the following functions with respect to x, using the result

d

dxaxn[ ] = naxn1.

(i) 15x6 , (ii)4

x3, (iii) x

3, (iv)

1

x4, (v) 4.

Find the gradient of the tangent to the curve y= 1

x4at the point 1

2, 16.


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30 ME 22A Unit 1

Differentiation of a Polynomial Function

Ify= f(x) +g(x),dy

dx= lim

h0

f x +h( ) +g x +h( ) f x( ) g x( )h

= limh0

f x +h( ) f x( )h

+ lim

h0

g x +h( ) g x( )h

wheredy

dxindicates the derivative ofywith respect to x.

Thereforedy

dx= f' x( ) +g' x( ) , i.e. we can differentiate a polynomial function term by term.

Example 1

Find the gradient of the tangent to the curve y=5x33x

2+2x 1 at the point (2, 31).

The gradient function is obtained by findingdy

dx=15x

26x +2.

Putting x = 2,dy

dx= 60 12 +2 = 50.

Example 2

Find the gradient of the curve y=4

x

1

x3at the point where x = 4.

y=4x

12 x

3, making

dy

dx=2x

32+3x

4=

2

x( )3+

3

x4.

Therefore, ifx = 4,dy

dx=

2

8+

3

256=

64

256+

3

256=

61

256.


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ME 22A Unit 1 31

y

x-b/2a

B M A

x1

p (x1,y1)

0

ACTIVITY 1.11(i) Find the gradient of the tangent to the curve y=3 2

x

at the point (1

2

, -

1).

(ii) Find the gradient of the tangent to the curve y=5 2x 4x2 at the pointwhere x = 1 .

Can you find the equation of this tangent?

To end this lesson (and Unit 1), we will look once again at the construction of the tangent

to the curve y=ax2+bx + c at the point (x1 , y1). If you remember, the technique was as

follows (this time we will take a > 0).

Fig. 16

IfB is the minimum point, we draw a line throughB parallel to the x-axis and a line

throughP(x1 , y1) parallel to the y-axis, the two lines meeting atA.Mis the mid-point of

AB andMPis the required tangent. Can you justify this method using your knowledge of

differentiation and a few algebraic skills?

Of course we know that, ify=ax2+bx + c, then

dy

dx= 2ax + b. The gradient when x = x1is

therefore 2ax1 + b. We need, then, to show that the gradient ofMP(in Fig. 1) is also

2ax1 + b. Be warned, however the following proof is a bit heavy!


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32 ME 22A Unit 1

Proof

As ax2+bx + c=a x

2+

bx

a+

c

a( ) =a x + b2a( )

2

+c

a b

2

4a

2

( )

=a x + b

2a[ ]2+ c b

2

4a( ), the minimum

value of the function is c b2

4a( ), this occurring when x = b

2a. The coordinates ofB are

therefore b2a

,c b2

4a( ) .

The length ofAB is therefore x1 b

2a( )

= x1 +

b

2a( ), and the length ofAM= 1

2x1 +

b

2a( )

.

Also, the length ofAPis y1 cb2

4a( )

=ax1

2+ bx1 + c c+

b2

4a( )= ax1

2+ bx1 +

b2

4a( ).

Therefore the gradient of MP =AP

AM=

ax12+ bx1 +

b2

4a( )

12

x1 +b2a( )

=

4a2x12+4abx1 +b

2

2ax1 + b=

2ax1 +b( )2

2ax1 +b

= 2ax1 +b, which is the required result.


33/34

ME 22A Unit 1 33

Exercise 1.4

1. Differentiate the following functions with respect to x, using the result

d

dxax

n( ) = naxn1.

(i) 3x3

1

x3, (ii)

4

x+3 , (iii)

2x3 3x2 ++1

x2,

(iv) 3x24( )

2

2. Find the gradient of the tangent to the curve y=4

xat the point (9, 12).

3. Find the equation of the tangent to the curve y=2x3 3x2 +5x 1 at the point (-1, - 11).

4. Find the equation of the tangent to the curve y=3 4x3

x

at the point where x = 1.

[The solutions will be given in the Appendix.]


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unit 1- the derivative

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