(A) Function of a function Suppose we want to differentiat (2x-1)3 . We could expand the bracket then differentiate term by term , but this is tedious ! We need a more direct method for expression of this kind . Now (2x-1)3 is a cubic function of the linear function (2x-1) , i.e. it is a function of a function . Other examples : (a) (x2 3 ) 3 is a cubic function of an quadratic function . (b) 41 x+ is a square root function of a quartic function . So how we differentiate the above function ????? this differentiation technique is called The Chain Rule . Definition of Chain Rule : Definitions : If y = f ( u ) and u = g ( x ) , then

.dy dy dudx du dx

= where dydu

is evaluated at u = g ( x ) .

Proof : Consider y = f(u) where u = u (x ) .

For a small change of x in x , there is a small change of u(x+ x ) u(x) = u in u and a small change of

y in y . (see figure 1)

Now , y y ux u x

= ( fraction multiplication)

As x 0 , u 0 also .

0 0 0

lim lim limx x x

y y ux u x

= Figure 1

.dy dy dudx du dx

= -------- (1)

In general , if y = f (v ) , v = g(u) , u=h(x) , then . .dy dy dv dudx dv du dx

=

In (1) , if let u = y , we can prove 1dy dxdxdy

=

In Leibniz Notation , the chain rule look like like a fake cancellation :

.dy dy dudx du dx

=

x x

u

x x+

u

u+

u u=u(x)

y u x

y

y=f(u)

u

y+

X

Dummy variable u

y

1

Example 1: Given y = 2u3 + 3 , u = 3x2 + 2x -1 , find dydx

. Example 2: Given y = u4 , u = x 2 2x , find dydx

.

[solution:] y = 2u3 + 3 and u = 3x2 + 2x -1 [8x3(x-2)3(x-1)]

Then dydu

= 6u2 and dudx

= 6x + 2

.dy dy dudx du dx

= = 6u2 (6x + 2 )

= 6(3x2 + 2x -1)2(6x+2) = 12(3x2 + 2x -1)2 (3x + 1 )

Example 3: Differentiate y = (3x2 + 2)8 Example 4 : Differentiate y = 23 5x + . [2

33 5

xx +

]

[solution:] let u = 3x2 + 2 , and so y = u8

Then dudx

= 6x and 78dy udu

=

By the chain rule ,

.dy dy dudx du dx

=

= 8u7 ( 6x ) = 48 x (3x2 + 2) 7

Example 5: Given x = t- 3t2 , y = 5t4 , find dydx

Example 6 : Given parametric equation 2

3

23

x t ty t t

=

= ,

in terms of t . find the gradient of the tangent at t= 13 .

[solution:] x = t 3t2 3 2dx tdt= [1]

y = 5t4 dydt

= 20 t3

.dy dy dtdx dt dx

=

= 1.dy dxdxdt

=20 t3 . 13 2t

= 320

3 2t

t

From the above , we can use Outside-Inside Rule to make it easily 1( )n nd duu nudx dx

=

E.g. 2

2 3(3 1)d xdx

+ = 2 12 232 (3 1) . (3 1)

3dx xdx

+ + =

12 32 (3 1) (6 )

3x x

+

Outside inside left alone differentiate inside

= 3 2

43 1

xx +

2

Example 7: Differentiate w.r.t x : (a) y= ( 1- 3x3 )5 (b) y = 5( )baxx

+ , where a,b are constants (c) y = 1 , 0x x+ (d) y = 5 23( ( 2) 8)x + Example 8: If x = (1-3t2)3 , y = 1

2 1t +, find Example 9 : If y =

2

11 2t+

and x = 1- t 3 , find dydx

in the value of dy

dxwhen t = 1 . [ 1

216 3] terms of t . [ 3

2 2

2

3 (1 2 )t t+]

Some Important Algebraic Identities 1. x n y n = ( x y ) (x n-1 + xn-2 y + x n-3 y2 + ..+y n-1 ) , n is positive integer . 2. x n y n = ( x + y ) (x n-1 - xn-2 y + x n-3 y2 - ..- y n-1 ) , n is even 3. x n + y n = ( x + y ) (x n-1 - xn-2 y + x n-3 y2 - ..- y n-1 ) , n is odd

3

Example 10 : Find 3 32 21 12 2

limx a

x a

x a

. Example 11 : Find 4 4lim

x a

x ax a

. [solution:] 3 32 21 1

2 2

limx a

x a

x a

[solution:] 4 4lim

x a

x ax a

= 4 41 1limx a

x ax a

Let 12x T= , 12a = t = As x a , T t Then 3 3 2 2( )( )lim lim( )T t T t

T t T t T Tt tT t T t + +

=

= 2 2lim( )T t

T Tt t

+ + = t2 +( t ) t + t2 = 3t 2 = 1 223( )a = 3a Example 12 : Find the following limit : (a) 9 93 3limx a x ax a [ 3a4 ] (b) 2 23 32 2limx a x ax a [ 4313 a ] (B) Rate of change of Connected Variables Rate of change includes the procedure of calculating rates of change of the given time variable function with respect to change in the input .

Rate of change of y = dydt

Rate of change of x = dxdt

If two variables x and y are connected by the equation y = f(x) ,

dy dy dxdt dx dt=

Notes : If x change at the rate of 5 cm/s 5dxdt=

Rate of Change

dAdt= Rate Of Change

of Area

dVdt

=Rate Of Change

of Volume

drdt= Rate Of Change

of radius

dhdt= Rate Of

Change of height

dldt= Rate Of Change

of length

4

Example 13: The radius , r cm , of a circular hole is increasing at the rate of 0.5 cm/s . Find the rate at which the area A of the hole is increasing when r = 5cm . [Solution] Given drdt

= 0.5 cm/s A = r 2 dA dA drdt dr dt= = 2 r dr

dt when r = 5 , dA

dt=2(5)(0.5) = 5 cm2 / s .

The area A is increasing at the rate of 5 cm2 /s . Example 14: The radius r cm of sphere is increasing at a constant rate of 2cm s-1 . Find , in terms of , the rate at which the volume is increasing at the instant when the volume is 36 cm3 . [ 72 cm3 s-1 ] Example 15: Water runs into a conical tank at the rate of 10cm3/min . The tank stands point down and has a height of 15cm and a base radius 5cm . How fast is the water level rising when the water is 4cm deep ? Find the rate of increase of the area of the horizontal surface of the liquid . [Solution:] Let h = depth of water in tank at time t r = radius of the surface of the water at time t V = volume of the water in the tank at time t Given dV

dt= 9cm3 /s , h = 6cm

Review Question : 1. The volume of a sphere is increasing at the rate of 2cm3/s . Find , in terms of , the rate of increase of its radius when the radius of the sphere is 10cm . [ 2007 UEC Q10(a) ; 1200

cm/s ]

hcm 15cm

5 cm

r cm

5

Example 16: A ladder 4m long is leaning against the vertical wall of a house . If the base of the ladder is pulled horizontally away from the house at a rate of 0.7 metres per second , how fast is the top of the ladder sliding down that wall at the instant when it is 2m from the ground ? [Solution:] Let x be the distance of the top of the ladder from the ground . y be the distance of between the ladder and the house . By Pythagorean Theorem , when x = 2 , Given dydt

= 0.7 dxdt= Hence the top of ladder is sliding down the house at ____________ m/s .

Example 17: A man of height 1.9m walks at the rate of 3.2 km/h directly away from a street lamp which is 6m above the ground . At what rate is the length of his shadow changing in metres per second ? [0.412 m/s] Example 18: The length of each side of a cube increases at a rate of 0.4 cm/s . Calculate the rate of change in its total surface area when its volume is 125cm3 . [24 cm2 s-1 ] ======================== The End =============================

x

y

ladder

Wall

Ground

6