THE CHAIN RULE (CASE 1)

1. THE CHAIN RULE (CASE 1) Suppose that z=f (x, y) is a differentiable function of x and y, where x=g (t) and y=h (t) and are both differentiable functions of t. Then z is a differentiable function of t and

dt

dy

y

f

dt

dx

x

f

dt

dz

dt

dy

y

z

dt

dx

x

z

dt

dz

Question No. 1.Use THE CHAIN RULE to find the derivative of w = xy with respect to t along the path x = cost and y = sint. What is the derivative’s value at t = Pie/2.

).........(idt

dy

y

w

dt

dx

x

w

dt

dw

Solution:We Use THE CHAIN RULE to find the derivative of w = xy with respect to t along the path x = cost and y = sint as follows:

xy

xy

dy

dwandy

x

xy

dx

dw

)()(

tdt

td

dt

dyandt

dt

td

dt

dxcos

sinsin

cos

Solution continued……….

ttt

tttt

txty

dt

dy

y

w

dt

dx

x

w

dt

dw

ieqnfromThen

2cossincos

).(coscos)sin.(sin

)(cos)sin(

)(

22

1cos

2.2cos

2/

tdt

dw

partFurther

2. THE CHAIN RULE (CASE 2) Suppose that w=f (x, y, z) is a differentiable function of x, y and z and are all differentiable functions of t. Then w is a differentiable function of t and

dt

dz

z

f

dt

dy

y

f

dt

dx

x

f

dt

dw

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dw

Question No. 2.Use THE CHAIN RULE to find the derivative of w = xy + z with respect to t along the path x = cost , y = sint and z = t. What is the derivative’s value at t = 0.

).........(idt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dw

Solution:We Use THE CHAIN RULE to find the derivative of w = xy + z with respect to t along the path x = cost, y = sint and z = t as follows:

1)()(

,)(

z

zxy

dz

dwandx

y

zxy

dy

dwandy

x

zxy

dx

dw

1cossin

,sincos

dt

dt

dt

dzandt

dt

td

dt

dyt

dt

td

dt

dx

Solution continued……….

ttt

tttt

txty

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dw

ieqnfromThen

2cos11sincos

1).(coscos)sin.(sin

1.1)(cos)sin(

)(

22

211

0.2cos10

tdt

dw

partFurther

3. THE CHAIN RULE (CASE 3) Suppose that z=f (x, y) is a differentiable function of x and y, where x=g (s, t) and y=h (s, t) are differentiable functions of s and t. Then

ds

dy

y

z

ds

dx

x

z

dx

dz

dt

dy

y

z

dt

dx

x

z

dt

dz

4. THE CHAIN RULE (GENERAL VERSION) Suppose that u is a differentiable function of the n variables x1, x2,‧‧‧,xn and each xj is a differentiable function of the m variables t1, t2,‧‧‧,tm Then u is a function of t1, t2,‧‧‧, tm and

for each i=1,2,‧‧‧,m.

i

n

niii t

x

x

u‧‧‧

dt

x

x

u

dt

dx

x

u

t

u

2

2

1

1

F (x, y)=0. Since both x and y are functions of x, we obtain

But dx /dx=1, so if ∂F/∂y≠0 we solve for dy/dx and obtain

0

dx

dy

y

F

dx

dx

x

F

y

x

F

F

yFxF

dx

dy

F (x, y, z)=0

But and

so this equation becomes

If ∂F/∂z≠0 ,we solve for ∂z/∂x and obtain the first formula in Equations 7. The formula for ∂z/∂y is obtained in a similar manner.

0

x

z

z

F

dx

dy

y

F

dx

dx

x

F

1)( x

x 1)( y

x

0

x

z

z

F

x

F

zFxF

dx

dz

zFyF

dy

dz

1. DEFINITION A function of two variables has a local maximum at (a, b) if f (x, y) ≤ f (a, b) when (x, y) is near (a, b). [This means thatf (x, y) ≤ f (a, b) for all points (x, y) in some disk with center (a, b).] The number f (a, b) iscalled a local maximum value. If f (x, y) ≥ f (a, b) when (x, y) is near (a, b), then f (a, b) is a local minimum value.

2. THEOREM If f has a local maximum or minimum at (a, b) and the first order partial derivatives of f exist there, then fx(a, b)=1 and fy(a, b)=0.

A point (a, b) is called a critical point (or stationary point) of f if fx (a, b)=0 and fy (a, b)=0, or if one of these partial derivatives does not exist.

3. SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are continuous on a disk with center (a, b) , and suppose that

fx (a, b) and fy (a, b)=0 [that is, (a, b) is a critical point of f]. Let

(a)If D>0 and fxx (a, b)>0 , then f (a, b) is a local minimum.(b)If D>0 and fxx (a, b)<0, then f (a, b) is a local

maximum.(c) If D<0, then f (a, b) is not a local maximum or

minimum.

2)],([),(),(),( bafbafbafbaDD xyyyxx

NOTE 1 In case (c) the point (a, b) is called a saddle point of f and the graph of f crosses its tangent plane at (a, b).NOTE 2 If D=0, the test gives no information: f could have a local maximum or local minimum at (a, b), or (a, b) could be a saddle point of f.NOTE 3 To remember the formula for D it’s helpful to write it as a determinant:

2)( xyyyxxyyyx

xyxyfff

ff

ffD

1444 xyyxz

4. EXTREME VALUE THEOREM FOR FUNCTIONS OF TWO VARIABLES If f iscontinuous on a closed, bounded set D in R2, then f attains an absolute maximum value f(x1,y1) and an absolute minimum value f(x2,y2) at some points (x1,y1) and (x2,y2) in D.

5. To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D:1. Find the values of f at the critical points of in D.2. Find the extreme values of f on the boundary of D.3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.