PARTIAL DERIVATIVES
Prepared by : MISS RAHIMAH JUSOH @
AWANG
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Objectives :
• Identify domain and range of function of
two and three variables.
• Sketch graphs and level of curves of
functions of two and three variables
• Compute first and second partial
derivatives.
Definition :
A function of two variables is a rule f that
assigns to each ordered pair (x,y)
in a set D a unique number z = f (x,y).
The set D is called the domain of the
function, and the corresponding values
of z = f (x,y) constitute the range of f .
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EXAMPLE 1
Find the domains and range of the following functions
and evaluate f at the given points.
a)
x  y 1
f  x, y  
;
x 1
Eva1uate f  3, 2 


6
Answer : D   x, y  x  y  1  0, x  1 , f  3, 2  
2
z  f  x, y  , range is  z    z  
b)
xy
f  x, y  
;
x  2y
Eva1uate f  2,3 , f  2,1
Continue…
Find the domains and range of the following functions
and evaluate f at the given points.
c)
f  x, y   25  x 2  y 2 ;
Eva1uate f  2,3 , f  7, 4 


Answer : D  x, y  x 2  y 2  25 ,
z  f  x, y  , range is  z 0  z  5
d)
f  x, y   3ln  y 2  x 
Eva1uate f  3, 2 
A set of points where f is a constant is
called a level curve. A set of level
curves is called contour map.
Figure 13.1.4 (p. 909)
Figure 13.1.8 (p. 911)
Figure 13.1.10 (p. 912)
Table 13.1.2a (p. 913)
Table 13.1.2b (p. 913)
Table 13.1.2c (p. 913)
Table 13.1.2d (p. 913)
Table 13.1.2e (p. 913)
Table 13.1.2f (p. 913)
f  x, y   9  x 2  y 2
or
z 2  x 2  y 2  32
x y 
f  x, y   sin 

 2 
2
2
EXAMPLE 2
Sketch the level curves of the function
f  x, y   9  x  y
2
for k  0,1, 2,3
2
f  x, y   9  x  y
2
2
Limits Along Curves
For a function one variable there two one-sided limits at a point x 0
namely
lim f ( x) and lim f ( x)
x  x0 
x  x0 
reflecting the fact that there are only two directions from which x
can approach x 0
Function of
2 variables
Function of
3 variables
lim
( x , y )( x0 , y0 )
f ( x, y)  lim f ( x(t ), y(t ))
t t0
( alongC )
lim
( x , y , z )( x0 , y0 , z0 )
f ( x, y, z)  lim f ( x(t ), y(t ), z (t ))
t t0
( alongC )
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EXAMPLE 3
xy
Find the limit of f ( x, y )   2
along
2
x y
(a) the x - axis
(b) the y - axis
(c) the line y  x
(d ) the line y   x
(e) the parabola y  x 2
The process of differentiating a function of
several variables with respect to one of its
variables while keeping the other variable(s)
fixed is called partial differentiation, and the
resulting derivative is a partial derivative of the
function.
The derivative of a function of a single variable
f is defined to be the limit of difference
quotient, namely,
f  x  x   f  x 
f   x   lim
x 0
x
Partial derivatives with respect to x or y are
defined similarly.
If z  f  x, y  , then the partial derivatives of f with
respect to x and y are the functions f x and f y ,
respectively, defined by
f  x  x, y   f  x, y  and
f  x, y   lim
x
x 0
x
f  x, y  y   f  x, y 
f y  x, y   lim
y  0
y
provided the limits
exists.
We can interpret partial derivatives as rates of change.
If z  f ( x, y) , then
z / x
represents the rate of
change of z with respect to x when y is fixed. Similarly,
z / y
is the rate of change of z with respect to y when x is
fixed.
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EXAMPLE 4
If f ( x, y )  4  x 2  2 y 2 , find f x (1,1) and f y (1,1).
Solution :
f x ( x, y )  2 x
f y ( x, y )  4 y
f x (1,1)  2
f y (1,1)  4
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EXAMPLE 5
If
 x 
f
, calculate f and
f ( x, y)  sin
.
y
x y
x
Implicit Differentiation (i)
Theorem : If F(x,y )  0 implicitly defines a differenti able
function y as a function of one variable x, then
Fx ( x, y )
dy

dx
Fy ( x, y )
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EXAMPLE 6
dy
Find
, if y as a function of x is implicitly defined by
dx
y 5  3 y-4 x 3-5 x  1
Solution :
We have y5  3 y-4 x 3-5 x  1 ,
so F ( x, y )  y 5  3 y-4 x 3-5 x  1
Thus
Fx ( x, y )
dy

dx
Fy ( x, y )
(12 x 2  5) 12 x 2  5


4
5y  3
5y4  3
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Implicit Differentiation (ii)
If z = F (x, y, z) then
Fy
Fx and z
z


x
Fz
y
Fz
EXAMPLE 7
Find z and z if z is defined implicitly as a function of
x
y
x and y by the equation
(a) x 3  y 3  z 3  6 xyz  1
(b) x 2 z 2  xy 2  z 3  4 yz  5  0
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
 f
2 f
f xx 
fx  ( )  2
x
x x
x

 f
2 f
f yy 
fy  ( )  2
y
y y
y

 f
 f
f yx 
fy  ( ) 
x
x y
xy
2

 f
 f
f xy 
fx  ( ) 
y
y x
yx
2
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EXAMPLE 8
Find f xx , f xy, f yx , f yy if f ( x, y )  x 2 y 3  x 4 y
The first derivative of f are

f x  ( x 2 y 3  x 4 y )  2 xy 3  4 x 3 y ,
x
The second derivative of f are
fy 
 2 3
( x y  x 4 y)  3x 2 y 2  x 4
y


f xx  ( f x )  (2 xy 3  4 x 3 y )  2 y 3  12 x 2 y
x
x


f xy  ( f x )  (2 xy 3  4 x 3 y )  6 xy 2  4 x 3
y
y


f yx  ( f y )  (3 x 2 y 2  x 4 )  6 xy 2  4 x 3
x
x


f yy  ( f y )  (3 x 2 y 2  x 4 )  6 x 2 y
y
y
EXAMPLE 9
3
2 3
2
Find the second partial derivatives of f ( x, y )  x  x y  2 y
EXAMPLE 10
Find f xx, f xy, f yx, f yy if given f ( x, y)  ( x  y)e xy
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Theorem : If z  f (u,v ), u  g ( x,y ) and v  h( x,y ) with partial
derivative s of f , g and h exist, then
z z u z v


,
x u x v x
z z u z v


,
y u y v y
The three diagram for function of two variables
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EXAMPLE 11
Suppose z  u 3  v 3 , and u  xy 2 , v  x 2 sin y
Use a chain rule z x and z y .
Solution :
From z  u 3  v 2 , we obtain
z
z
2
 3u
and
 2v
u
v
Similarly from u  xy 2 and v  x 2 sin y,
u
u
v
 y2 ,
 2 xy,
 2 x sin y ,
x
y
x
Hence we obtain
z z u z v


 3u 2 ( y 2 )  2v(2 x sin y )
x u x v x
 3u 2 y 2  4vx sin y
v
 x 2 cosy
y
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continue...
z z u z v


 3u 2 (2 xy )  2v( x 2 cos y )
y u y v y
 6u 2 xy  2vx2 cos y
Substituting u  xy 2 and v  x 2 sin y, weobtain
z
 3u 2 y 2  4vx sin y
x
 3( xy 2 ) 2 y 2  4( x 2 sin y ) x sin y
 3x 2 y 6  4 x 3 sin 2 y
z
 6u 2 xy  2vx2 cos y
y
 6( xy 2 ) 2 xy  2( x 2 sin y ) x 2 cos y
 6 x 3 y 5  x 4 sin 2 y
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EXAMPLE 12
2
4
z

x
y

3
xy
, where x  sin 2t and y  cost ,
If
find dz when t = 0 .
dt
Solution :
The Chain Rule gives
dz z dx z dy


dt x dt y dt
2
4
we calculate the derivatives, since z  x y  3 xy , we get
z
 2 xy  3 y 4
x
z
 x 2  12 xy 3
y
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Then from x  sin 2t and y  cost , we get
dx
 2 cos 2t
dt
dy
  sin t
dt
So,
dz z dx z dy


 (2 xy  3 y 4 )( 2 cos 2t )  ( x 2  12 xy 3 )(  sin t )
dt x dt y dt
 (2 sin 2t cos t  3 cos 4 t )( 2 cos 2t )  (sin 2 2t  12 sin 2t cos 3 t )(  sin t )
Therefore
z
 (0  3)( 2)  (0  0)( 0)  6
t t 0
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Chain rule can be applied to compositefunctions of
any number of variables, and tree diagrams can be
constructed to help formulate these rules.
EXAMPLE 13
Use the chain rule to find dw dt if
w  x 2  3 yz with x  3t 2  4, y  2t  1, and z  t 3
Solution :
x
w
t
y
t
z
t
dw w dx w dy w dz



dt x dt y dt z dt
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From the tree diagram, we obtain
dw w dx w dy w dz



dt x dt y dt z dt
 2 x(6t )  3z (2)  3 y (3t 2 )
 12tx  6 z  9t 2 y
 12t (3t 2  4)  6(t 3 )  9t 2 (2t  1)
 3t (20t 2 -3t  16)
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EXAMPLE 14
The pressure P (in kilopascals), volume V (in liters), and
temperature T (in kelvins) of a mole of an ideal gas are related by
the equation PV  8.31T . Find the rate at which the pressure is
changing when the temperature is 300 K and increasing at a rate
0.1 K/s and the volume is 100 L and increasing at a rate of 0.2 L/s.
Solution :
If t represent the time elapsed in seconds, then at the given
instant we have
T  300,
dT
dV
 0.1, V  100,
 0.2.
dt
dt
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The chain rule gives
dP P dT P dV 8.31 dT 8.31T dV




dt
t dt V dt
V dt
V 2 dt
8.31
8.31(300)

(0.1) 
(0.2)
2
100
100
 0.04155
The pressure is decreasing at a rate of about 0.042 kPa/s.
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Figure 13.6.2 (p. 960)
Theorem 13.6.3 (p. 961)
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Figure 13.6.5 (p. 964)
Theorem 13.6.5 (p. 964)
Objective :
•
•
Compare absolute extrema and local extrema.
Locate critical points and determine its
classification using second partial derivatives test.
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Definitions 13.8.1 and 13.8.2 (p. 977)
Figure 13.8.2 (p. 977)
 Critical point (or stationary point) of graph z  f ( x,y) has
three possibilit ies :
- maximum point, minimum point or saddle point.
The wordextremum refer toeither maximum or minimum
values of a function.
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Definition
Suppose z  f ( x,y) is a function of two variables
defines in a domain D, and that (a, b) in D
(i) f (a,b) is a local maximum of the function f , if there exists a
region R  D, such that f (a,b)  f ( x,y) for every ( x,y) in R.
(ii) f (a,b) is a local minimum of the function f , if there exists a
region R  D, such that f (a,b)  f ( x,y) for every ( x,y) in R.
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Point (a,b,f (a,b)) is a Local Maximum
Point (a,b,f (a,b)) is a Local Minimum
Point (a,b,f (a,b)) is a saddle point
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A pair (a,b) such that
f x (a, b)  0 and f (a, b)  0 is
y
called a critical point or stationary point. To find out
whether a critical point will give f (x, y) a local maximum or
a local minimum, or will give a saddle point, we use
theorem : Second Derivative Test.
Theorem: SecondDerivativeTest
Suppose that f ( x,y) is a function of two variables that has
continuous second partial derivatives on a region R and (a, b)
is a critical point in R. Let
G(a,b) 
f xx (a, b)
f xy (a, b)
f xy (a, b)
f yy (a, b)
 f xx (a, b) f yy (a, b)  [ f xy (a, b)]2
Then
(i)
f (a, b) is a local maximum if G(a,b)  0 and f xx (a, b)  0
(ii)
f (a, b) is a local minimum if G(a,b)  0 and f xx (a, b)  0
(iii) (a, b, f (a, b)) is a saddle point if G(a,b)  0
(iv) No conclusions can be made if G(a,b)  0
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EXAMPLE 15
Find the critical points of the following functions and
determine whether f (x,y) at that point is a local maximum or a
local minimum, or value at a saddle point.
(i)
(ii )
(iii )
y3
2
f ( x, y ) 
x y
3
f ( x, y )  x 4  16 xy  y 4
1 1 1
f ( x, y )   
x y xy
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3
y
Solution :
(i) f ( x, y ) 
 x2  y
3
Step 1 : Find the first partial derivatives
f x  2 x
, fy  y 2  1
Step 2 : Solve f x  0 and f y  0 simultaneously
-2 x  0
y 2 1  0

x0

y  1
Step 3 : Find the second derivative and G( x,y)
f xx  2
f xy  0
f yy  2 y
 G( x,y)  f xx f yy  f xy2  2(2 y )  0  4 y
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Step 4 : Test thepoint (0,1) and (0,-1)
At (0,1) G(0,1)  - 4(1)  4  0
1
2
f (0,1)   0  1  
3
3
2
Therefore,(0,1,- ) is a saddle point
3
At (0,-1)
G(0,-1)  -4(1)  4  0
f xx (0,1)  2  0
( 1) 3
2
f (0,1) 
 0 1 
3
3
2
Therefore, (0,-1, ) is a maximum point
3
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Clearly finding the local extremum is not enough
for most practicalapplicatio ns. It is more useful
to find the extremum value of f ( x,y) on
only given region.
Definition
Absolute extremum for a function is the
extremum value on any given region.
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Theorem
Suppose f ( x,y) is continuous in a region R.
(i) If f ( x,y) assumesonly a local maximum and there is no
local minimum and no saddle point in the region, then
thelocal maximum is the absolute maximum in the region.
(ii) If f ( x,y) assumesonly a local minimum and there is no
local maximum and no saddle point in the region, then
thelocal minimum is the absolute minimum in the region.
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EXAMPLE 16
Find the absolute extremumof f ( x, y)  x3  y 3  6 xy
in the region R  ( x, y) : 0  x  ,0  y  .
Solution :
Step 1 : Find critical points.
Find ( x,y) such that f x ( x, y )  0 and f y ( x, y )  0 simultaneously.
Thus we have
3 x 2  6 y  0 and 3 y 2  6 x  0
Solving the equations simultaneously, we obtain critical points (0,0)
and (2,2).Since (0,0) is outside the region R, we only test the critical
point (2,2)
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Step 2 : Compute f xx (2,2), f xy (2,2) and f yy (2,2)
f xx ( x,y)  6 x
thus f xx (2,2)  12
f xy ( x,y)  6
thus f xy (2,2)  6
f yy ( x,y)  6 y
thus f xy (2,2)  12
Step 3 : Compute G(2,2) and classify the critical point.
G(2,2)  12(12 )-(-6) 2  108  0 and f xx (2,2)  12  0
Therefore
f (2,2)  23  23  6(2)( 2)  8
is a local minimum.
Since there is no local maximum in R,
therefore the local minimum f (2,2)  -8
is the absolute minimum in R.
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“Just believe in yourself and work hard, no
matter what obstacles or hardships come in
your way. You will definitely reach your final
destination.”
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