DIFFERENTIATION - Web viewIf B is another point on the curve, fairly close to A, then the gradient of the chord AB gives an approximate value for the gradient of the tangent at A

UNIT 8: PURE MATHEMATICS 1 – DIFFERENTIATION

DIFFERENTIATION

Candidates should able to:

Understand the idea of the gradient of a curve, and use the notations

f ' ( x ) , f ' ' ( x ) , dydx and

d2 yd x2 ;

Use the derivative of xn (for any rational n), together with constant multiples, sums,

differences of functions, and of composite functions using the chain rule; Apply differentiation to gradients, tangents and normal, increasing and decreasing

functions and rates of change (including connected rates of change); Locate stationary points, and use information about stationary points in sketching

graphs (the ability to distinguish between maximum points and minimum points is required, but identification of points of inflexion is not included).

1. Gradient of A CurveThe gradient of a curve is not constant; it has different values at different points on the curve. In other words, the gradient varies as the point changes and it is equal to the gradient of a tangent to the curve at that point.

Consider the problem of finding the gradient of the curve at a pt A(x, y).If B is another point on the curve, fairly close to A, then the gradient of the chord AB gives an approximate value for the gradient of the tangent at A. As B gets nearer to A, the chord AB gets closer to the tangent at A, so the approximation becomes more accurate.So we say,

As B→ A,the gradient of chord AB → the gradient of the tangent at A.

Or limB→ A

(gradient if chord AB )=gradient of the tangent at A .

∴ The gradient of a curve at A(x, y) = Gradient of the tangent to the curve at A(x, y).

We now find the general expression for the gradient of a curve at any point using the process called differentiation. The general expression for the gradient of a curve y=f (x ) is itself a

function so it is called the gradient function. For example the curve y=x2, the gradient

function is 2 x . We shall use the notation dydx

to denote the gradient function.

Since the gradient function is derived from the given function, it is also called derived function or the derivative or differential coefficient of y with respect to x.

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Different notations are used to represent gradient function depending on the given question as shown below.

Given

y=2x4, find dydx

Given

f (x)=2x4, find f '

Given y=2x4, find y '

Evaluate ddx

(2 x4)

Differentiate

2 x4, w.r.t. x

y=2x4dydx

=2 (4 ) x3

¿8 x2

f (x)=2x4

f ' (x)=2 (4 ) x3

¿8 x2

y=2x4

y '=2 (4 ) x3

¿8 x2

ddx

(2 x4)=2 (4 ) x3

¿8 x2

ddx

(2 x4)=2 (4 ) x3

¿8 x2

Additional information (taken from SMSA note)

The derivative dydx

tells us the rate of change of one quantity compared to another at a

particular instant or point (so we call it “instantaneous rate of change”.)

The derivative tells us:1. The rate of change of one quantity compared to another 2. The slope of a tangent to a curve at any point

Applications include:1 Temperature change at a particular time2 Velocity of a falling object at a particular time 3 Current through a circuit at a particular time 4 Variation in stockmarket prices at a particular time5 Population growth at a particular time 6 Temperature increase as density increases in a gas

2. Basic rule to differentiate algebraic functions Given a function y=xn, we can use the following rule to find its derivative.

For example to find the derivative of 5 x3,

2.1 Techniques of DifferentiationAlgebraic functions can be presented in many forms. The following table shows the techniques to find the derivative of all possible forms of algebraic functions.

c is a constant nth power of x Constant product Derivative of a sum (or a difference)

dcdx

=0 dd x

xn=nxn−1 ddx

(cy )=c ddx

( y) d (u±v)dx

=dudx

± dvdx

2.1.1 The derivative of a constant is 0

y=k ,k is aconstant

ddx

(k )=0 If y=7 ,dydx

=¿

If y=−5,dydx

=¿

If y=c ,c i saconstantdydx

=¿

2.1.2 Power rule or nth power of x

e.g. 1:

If y=x4 , find dydx

.

e.g. 2:

Find ddx

(x12 )

e.g. 3: Differentiate,w.r.t. x: f ( x )= 3√x

e.g. 4: Evaluate

ddx

(x−12 )

dydx

=4 ( x4−1 )

¿4 x3

2.1.3 Differentiation with constant multiples or Constant producte.g. 5: Differentiate w.r.t. x:

y=3 x4 e.g. 6: Find ddx

(3 x12) e.g. 7: Given f ( x )=1

2x−2

,

find f ' .

2

1st meaning 2nd meaningdydx

gives the value of gradient at any point, P , on the curve, y = f (x)

gives the rate of change of y with respect to x


dydx

=3 (4 ) x4−1=12x3

2.1.4 Differentiating Sums or Differences

Any x term under root sign (e.g. √ x=x12) or in the denominator (e.g.

1x3

=x−3) must be

changed before we start to differentiate. Some common mistakes done by students are listed below. Study it and please try to avoid such mistakes.

× 12x4

=2 x−4

√ 12x4

= x−4

2

× 23√4 x

=2 (4 x )−13 =8 (x )

−13

√ 23√4 x

=2 (4 x )−13 =2 ( 4 )

−13 ( x )

−13

e.g. 8: Given y=3 x2+ 12 x

−3√ x+5, find

dydx

.

e.g. 9: Evaluate ddx

(x12+ 2

x+7)

Apply power rule and differentiate term by term.

y=3 x2+ x−1

2−x

13+5

dydx

=6 x− x−2

2−13x

−23 +0

Exercise 1: Differentiate each of the following functions w.r.t. x.

a) x5

b)1x4

c) √ xd) x−4

g) y=x5+ 1x4

−√ x+3

h) y=2x3−x23+ x

−32 −4

e) x23

f) x−32

i) f ( x )=3 x+ 13x4

−√x+ 12j)

f ( x )= 3x4

+5√x− 23 x

−2 23

k) y= 1√ x

− 24√3 x

+ √x2

+1

2.2 The second derivative d2 yd x2

If y is a function of x , then dydx

is also a function of x. Hence, we can differentiate dydx

w.r.t. x . This gives the second derivate which is written as d2 yd x2

(¿ f ' ' ). Or in another

words, we apply the rule of differentiation twice.

e.g. 10: Find d2 yd x2

or 2nd derivate for each of the following functions:

(a) y=2x3(b)

d2

d x2(√ x )=¿ (c) f ( x )= 1

2 x2

e.g. 11: Find d2 yd x2

and ( dydx )2

if

(a) y=2 x4−3 x3+x

x(b) y= 3√x (c) y= (x+2 )(2 x−1)

NOTE: d2 ydx2

is ______________________ to ( dydx )2

.

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2.3 Differentiation of composite functions (or function of a function) The function y=(x¿¿2+2)2 ¿ can be expanded as y=x4+4 x2+4 and so its first

derivative, dydx

=¿

Similarly, if we want to find the derivative of y= (2x−1 )5, we first expand it into

polynomial. This would be rather lengthy so we look for a better method. To do this, we

take (2 x−1 )5 as a composite or combined function or function of a function.

The function y= (2x−1 )5 can be built up from 2 simple functions; u=2x−1 and

then y=u5. We call u the core function. Now,

u is a function of x so dudx

=2 , y is a functionof u so dydu

=5u4.

To obtain dydx

from these two derivatives, we use a rule for the derivative of composite

functions.

dydx

=dydu

× dudx

,where y is a function of u∧u is a function of x .

Note: du cannot be cancelled on the right hand side as these are not fractions but derivatives. However the notation suggests the result and is easy to remember.

Then, dydx

=5u4×2=10u4=10 (2 x−1 )4.

OR

OR

Few examples of using Chain rule with different independent & dependent variables.dydx

=dydv

× dvdx

dVdr

=dVdA

× dAdr

dvdt

=dvds

× dsdt

dVdt

=dVdr

× drdt

dsdt

= dsdx

× dxdt

e.g. 12: Differentiate the following w.r.t. x

(a) y ¿(3 x2+2x )7 (b) y= 1(3x−2)

(c) y=√3 x2+5

e.g. 13: Find f '( x) and f '(2) if:

(a) f (x)¿(x3−x )32 (b) y=√3 x3−4 (c) y= 1

√ x+3xe.g. 14: Find the gradient of the tangent to a curve y=( x2−3 )5 at the point (−1 ,5).

Exercise 2: Find d2 ydx2

for each of the following functions.

(a) y=x5−4 x (c) y=x2+ 10x2

(e) y= 9√ x

+x

(b) y=x+ 13 x2

(d) y= (2x−1 )(x+3) (f)

y=x2(3 x2−2 x+3)

Exercise 3: Differentiation of Composite functions

1. For each of the following functions, apply chain rule directly to find dydx

.

(a) y=( x5+1 )7 (b) y=(6 x3−5 )−2 (c) y=( x2+3 x+1 )6

2. Find the gradient of the tangent to the curve y= 1√ x−1

at the point (−1 ,4 ) .

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CHAIN RULE

If y= (ax+b )n, then

dydx

=n (ax+b )n−1× ddx

(ax+b )

If y= [ f (x )]n, then

dydx

=n [ f (x )]n−1× f ' (x)


3. Equation of Tangent and Normal

Note:The line that touches the curve at point P is called the tangent at P. The line passing through P (the point of contact of the tangent with the curve) which is perpendicular to the tangent is called the normal to the curve at that point.

The gradient of the tangent to the curve y= f (x ) is given by mT=d ydx

=f '( x) at

x=x1. Hence, Equation of Tangent is

Equation of Normal is

Additional information:

e.g. 15: Find the equation of the tangent to y= 3√x at the point (8 ,2).

e.g. 16: The equation of the curve y=6−3 x−4 x2−x3. Find the gradient of the

tangent and of the normal to the curve at the point (−2 ,4 ) .

e.g. 17: Find the equation of the tangent and normal to the curve y=3 x2−4 x+5 at

the point where x=2.

e.g. 18: the equation of a curve is y=101+x2

. Find the equation of the normal to the

curve at the point where x=3.

5

y− y1=mT (x−x1)

y− y1=−1mT

(x−x1)

Graph shows the curve y= x2

2−4 x+6. Its

gradient function, dydx

=x−4 .

Interval Sign of

dydx

Interpretation Increasing or decreasing function

x<4 dydx

<0 y ↓ (decreases )as x ↑

( increases )umtilx=4

y is a decreasing function of x .

x>4 dydx

>0 y ↑ (increases )as x ↑

( increases ) ¿x=4

y is an increasing function of x .

When x=4, dydx

=0, this implies that y is

stationary at x=4.

Decreasing Function, dydx

<0Stationary Point,

dydx

=0 Increasing Function, d ydx

>0

On any stretch of the curve

y=f (x ), where dydx

<0,

then f ( x ) is a decreasing function, i.e. y decreases as x increases.

At any point where dydx

=0, f (x)

has a stationary value and is neither increasing nor decreasing. Such a point is a stationary point (turning point).

On any stretch of the curve

y=f (x ), where dydx

>0,

then f ( x ) is a increasing function, i.e. y increases as x increases.

decreasing

increasing

y↓as x ↑

y↑as x ↑ Given y = f(x).

Gradient function =

dydx

Gradient of the tangent at

any point, P =

dydx

Gradient of normal at any

point, P =

−1dydx


e.g. 19: The curve y=x2+3 x+4 passes through the points P(1 ,2) and Q(3 ,4 ).

The tangent to the curve at P intersect the normal to the curve at Q at point R . Calculate

the coordinates of R .

Exercise 4: Gradients of tangents and normals.

1. Find the equation of the tangent y=5 x2−7 x+4 at the point (2 ,10).

2. Find the equation of the normal to y=x4−4 x3 at the point x=12

.

3. Differentiate y=3 x2−2 x+5. Then, find the equation of the tangent and the

normal to the graph of y=3 x2−2 x+5 at the point x=1.

4. The equation of a curve is y=2x2−5 x+14. The normal to the curve at the point

(1 ,11) meets the curve again at the point P. Find the coordinates of P.

4. Increasing and Decreasing function

e.g. 20: For what range of values is the function y=x3−3 x2−9x+4,

(a) decreasing, (b) increasing?

e.g. 21:

(a) For what range of values of x is the function y=x+ 14 x

increasing?

(b) What are the coordinates of the stationary points?

e.g. 22: (J08/Q6)

The function f is such that f ( x )= (3x+2 )3−5 for x≥0. Obtain an expression for

f '( x) and hence explain why f is an increasing function.

e.g. 23: If y=x2−2 x+3, find the interval in which y is increasing.

e.g. 24: If y= x3

3−3x2+5 x+2, find the interval in which y is decreasing.

Exercise 5: Increasing and Decreasing Functions

1. Given f ( x )=7−3 x−x2, find f '( x) and the interval in which f ( x ) is increasing.

2. Given f ( x )=2x5−5 x4+10, find f '( x) and the interval in which f ( x ) is

decreasing.

3. Given f ( x )=2x3−18 x+5, find f '( x) and the interval in which f ( x ) is

increasing.

4. Given f ( x )=(2−√x )2, find f '( x). If y=f (x ), determine whether the function is

an increasing function or decreasing function on the interval 1<x<2.

5. Rate of Changedydx is the rate of change of y with respect to x .

Thus, dydx

=5 means y is increasing at 5 units for every increase of 1 unit of x .

Similarly, if Am2 (square metres) represents an area, and t s (seconds) represents a time

interval after a reference time, then dAdt

=3 means that the area is increasing at the rate

of 2m2/s. A negative value of dAdt

, for example, dAdt

=−1.5m2/ s, denotes a rate of

decrease of the area.

We now use differentiation to study the rate of change of certain variables with respect to time. We write rate as dt . Hence,

The rate of change of radius, r⇒ drdt

The rate of change of area, A⇒ dAdt

The rate of change of volume, V ⇒

The rate of change of distance, s⇒

6


e.g. 25:Statement form Equation form

1 Therate of increase of ywith respect ¿ t is 32 The rate of increase of r withrespect ¿ time , t is4 t3 Rate of decrease of A withrespect ¿t is104 The rate of increase of N is propotional ¿N

e.g. 26: The radius, r cm, of a circle at time t seconds is given by r=9 t−t3. At each of

the following instants, find the rate of change of the radius (w. r. t. t ) and state whether the radius is increasing or decreasing at these instants.(a) t=1

(b) t=2

e.g. 27: The volume, V litres of water in a tank after t seconds is given by V=5− 2t+1

(a) What is the initial volume of water?(b) Find the rate at which the volume of water is increasing when t=3.

5.1 Related Rates of Change Chain Rule

If x and y are related by the equation y= f (x ), then the rates of change dxdt

and dydt

are related by:

e.g. 28: Given that y=12

(5 x−x2 ) . If the rate of change of x is 2 units s−1, find the

rate of change of y when

(a) x=3 ,(b) x=8.

Useful formulas for some geometrical figures.Volumes Surface Area

Area

Additional information:

The rate of change of a variable x with respect to time t is dxdt

.

At the instant t=t 0 , dxdt

>0⇔x increasing as t increase, & dxdt

<0⇔x decreasing as t increases.

Steady rate: dxdt

=m, constant ⇔x=mt+c .

e.g. 29: Some oil is split onto a level surface and spreads out in the shape of a circle. The

radius r cm of the circle is increasing at the rate of 0.5 cm s−1. At what rate is the area of

the circle increasing when the radius is 5 cm? [5π cm s−1]

e.g. 30: A spherical balloon is being blown up so that its volume increases at a constant rate

of 1.5 cm3 s−1. Find the rate of increase of the radius when the volume of the balloon is 56

cm3. [0.0212 cm s−1]

e.g. 31: If the radius r of a sphere is increasing at 2cm s−1, find the rate at which the

volume of the sphere is increasing when radius is 3cm. (Leave answer in terms of π).

[0.0212 cm s−1]

7

dydt

=dydx

× dxdt

….ChainRule

140 cm

50 cm 50 cm

80 cm


e.g. 32: Water is poured into an inverted circular cone of base radius 5 cm and height 15 cm at the rate of 10 cm3/s. Calculate:(a) the rate of increase of the height of water level,(b) the rate of increase of the surface area of the water, when the water level is 4 cm high.

[48 /8π cm/s, 5 cm2 s−1]

e.g. 33: Given that, when r=4 cm, volume of sphere, V is increasing at a rate of 8 m3/s, find the rate of increase of surface area, A at this instant. [4 m2/s]

e.g. 34: Water is pumped into an empty trough which is 200cm long, at the rate of 33000 cm3/s. The uniform cross-section of the trough is an isosceles trapezium with the dimensions shown. Find the rate at which the depth of the water is increasing at the instant when this depth is 20 cm. [1.5 cm s−1] Exercise 6: Rate of Change1. The radius of a circular disc is increasing at a constant rate of 0.003 cm/s. Find the rate

at which the area is increasing when the radius is 20 cm. [0.377 cm2 s−1]

2. The volume, V cm3, of a cube at time t seconds is given by V=(4+ 13 t)3

. Find the

rate at which its volume is increasing at the instant when t =2. [21 79

cm3/s]

3. Given that, when r=5cm, volume of sphere, V is increasing at a rate of 10 m3/s, find the rate of increase of surface area, A at this instant. [4 m2/s]

4. The radius of a circle increases at the rate of 3cm/s. Find the rate of change of the area when (a) the radius is 5 cm, and (b) the area is 4π cm2. [30π cm2/s, 12π cm2/s]

6. Applications of Differentiation – Maximum and Minimum

Stationary points

A stationary point on a curve is defined as a point on the curve where the gradient ( dydx ) is

Zero.

If y=f (x ) represents the equation of a curve and P is a point on the curve such that dydx

at P is zero, then P is a stationary point.

Turning pointsIf the gradient of the function changes sign at the stationary point, then it is called a turning point. This point may be a maximum point or a minimum point.

Determination of Maximum and Minimum pointsTo determine the nature of the turning point, that is, if it is a maximum or minimum point we apply the first derivatives or second derivatives of the function.

6.1 First Derivative test Maximum and Minimum pointIf the gradient changes from positive to zero then to negative as x increases through a stationary, the point is a maximum point. If the gradient changes from negative to zero then to positive as x increases through a stationary point, the point is a minimum point.

Points of Inflection (or inflexion)If the gradient on a curve changes from positive to zero and then to positive again or negative to zero and then to negative again as x increases through a stationary point, the point is a point of horizontal inflection.

8


Local maximum and minimumP and Q are 2 stationary points on the curve y=f (x ) shown in the diagram. P is a maximum pt. However, this does not necessarily mean that the value of the y coordinate at P is the absolute greatest, as it can be seen that the value of the y coordinate at R, for example, is greater than that at P. Similarly, Q is a minimum pt only for parts of the curve near Q. Again this does not necessarily mean that the value of y coordinate at Q is absolute smallest as it can be seen , for example, that the value of the y coordinate at S is smaller than that at the point Q.

In short, maximum or minimum points are only relative to points in the neighborhood (given interval) of the specified turning points. Unless otherwise indicated, maximum and minimum points are understood to be relative or local maximum and minimum points which are not necessarily the absolute maximum or minimum points on the graph.

e.g. 35: Given the curve y=x3−2x2+x+2, find

(a) the stationary points on the curve,(b) determine if they are maximum or minimum points.

e.g. 36: Given the curve y=x3+1, find the stationary point on it and determine its

nature.

6.2 Second Derivative test To find a turning point

Given y=f ( x ) ,If its dydx

=0 and d2 ydx2

≠0, then S(a , f (a )) is a turning point. And

f (a) is the maximum or minimum value of y .

To test for a maximum point, minimum point or point of inflection

(a) If a point on a curve has dydx

=0 and d2 ydx2

<0⇒ the point is a maximum point.

(b) If a point on a curve has dydx

=0 and d2 ydx2

>0⇒ the point is a minimum point.

(c) If d2 ydx2

=0, no definite conclusion can be made at this stage. The point may be a

minimum or maximum point or point of inflection. To determine the type of stationary point, consider apply the first derivative test.

e.g. 37: Find the stationary values in the function y=2x3+3 x2−12x−12 and

determine if they are maximum or minimum values.

e.g. 38: Determine the value of x , where x>0, for which the curve y=x2+ 16x

has a

stationary point and determine whether it is a maximum or a minimum point.

6.3 Curve sketching – using stationary points and their natureSteps to follow:1. Find stationary points and its nature to start the sketch.2. Find x-intercepts and use these points to continue the sketch.3. Examine the behaviour of f (x) as x→±∞ to complete the sketch.

e.g. 39: Find the coordinates of any stationary points on the curve y=x4+2x3 and

distinguish between them. Hence sketch the curve.

e.g. 40: Sketch the curve of y=x3−6 x2+9 x .

e.g. 41: Sketch the curve y=x3−3 x+4.

Exercise 7: Maximum and Minimum

9


1. For the following, find the coordinates of any stationary points on the given curves and determine the nature of the stationary points.

(a) y=2x2−8 x(b) y=18 x−20−3 x2

(c) y=x3−x2−x+72. For the following, find the coordinates of any stationary points on the curve and

distinguish between them. Hence sketch the curve.

(a) y=5 x6−12 x5, (b) y=x4−4 x3

7. Maximum and Minimum values problems

e.g. 42: Two positive numbers x and y vary in such a way that xy=18. Another number

z is defined by z=2 x+ y. Find the values of x and y for which z has a stationary value

and show that this value of z is a minimum.

e.g. 43: a rectangular plot of land is to be enclosed by 100m of fencing. The fencing is only used on three sides of the rectangle. The fourth side is a straight river bank. What should the dimensions of the rectangle be, to give a maximum area? Show that this area is a maximum.

e.g. 44: the figure shows a wired cuboid. The length of the base is 2 x cm and the breadth

is xcm. The height is h cm. if the total length of wire is 720 cm, show that the volume of

the cuboid V cm3 is given by V=360 x2−6 x3.

Determine the stationary value of V as x varies and show that this is a maximum value.

e.g. 45: an equilateral triangle ABC of side 6 cm is to have a rectangle PQRS inscribed

in it as shown in the diagram. Show that the area of the triangle, A cm2, is given by

A=√3xx (6−x ), where PQ=x cm. calculate the value of x which A has a

stationary value. Show that this value of x makes A a maximum.

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Documents

DIFFERENTIATION - Web viewIf B is another point on the curve, fairly close to A, then the gradient of the chord AB gives an approximate value for the gradient of the tangent at A