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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.