Chapter 5(partial differentiation)

Engineering Mathematics 1

1. BMM 104: ENGINEERING MATHEMATICS I Page 1 of 8
CHAPTER 5: PARTIAL DERIVATIVES
Functions of n Independent Variables
Suppose D is a set of n-tuples of real numbers ( )
n21 x,...,x,x . A real valued function f
on D is a rule that assigns a unique (single) real number
( )
n21 x,...,x,xfw=
to each element in D. The set D is the function’s domain. The set of w-values taken on
by f is the function’s range. The symbol w is the dependent variable of f, and f is said to
be a function of the n independent variables
1x to nx. We also call the s'x
j the
function’s input variables and call w the function’s output variable.
Level Curve, Graph, surface of Functions of Two Variables
The set of points in the plane where a function ()y,xf has a constant value ()cy,xf=
is called a level curve of f. The set of all points ()( )y,xf,y,x in space, for ()y,x in
the domain of f, is called the graph of f. The graph of f is also called the surface
( )y,xfz= .
Functions of Three Variables
The set of points ( )z,y,x in space where a function of three independent variables has a
constant value ( )cz,y,xf = is called a level surface of f.
Example: Attend lecture.
Partial Derivatives of a Function of Two Variables
Definition:Partial Derivative with Respect to x
The partial derivative of ( )y,xf with respect to x at the point ( )
00
y,x is
( )
( ) ( )
,
h
y,xfy,hxf
lim
x
f
0000
0h
y,x
00
-+
=
¶
¶
®
provided the limit exists.
Definition:Partial Derivative with Respect to y

2. BMM 104: ENGINEERING MATHEMATICS I Page 2 of 8
The partial derivative of ( )y,xf with respect to y at the point ( )
00y,x is
( )
( )
( ) ( )
,
h
y,xfhy,xf
limy,xf
dy
d
y
f
0000
0h
yy0
y,x
0
00
-+
==
¶
¶
®
=
provided the limit exists.
Example:
1. Find the values of
x
f
¶
¶
and
y
f
¶
¶
at the point ( )5,4- if
( ) 1yxy3xy,xf
2
-++= .
2. Find
x
f
¶
¶
if () xysinyy,xf= .
3. Find xf and yf if ()
xcosy
y2
y,xf
+
= .
Functions of More Than Two Variables
Example:
1. Let ( )
32
zxyz,y,xf = . Find
x
f
¶
¶
,
y
f
¶
¶
and
z
f
¶
¶
at ( )1,2,1-- .
2. Let
( )
z
y
2
exz,y,xg =
. Find xg,yg and
zg.
Second-Order Partial Derivatives
When we differentiate a function ()y,xf twice, we produce its second-order
derivatives.
These derivatives are usually denoted by
2
2
x
f
¶
¶
“ d squared fdx squared “ orxxf“f sub xx “
2
2
y
f
¶
¶
“ d squared fdy squared “ oryyf“f sub yy “
2
2
x
f
¶
¶
“ d squared fdx squared “ orxxf“f sub xx “
yx
f
2
¶¶
¶
“ d squared fdxdy squared “ or yxf“f sub yx “

3. BMM 104: ENGINEERING MATHEMATICS I Page 3 of 8
xy
f
2
¶¶
¶
“ d squared fdydx squared “ or xyf“f sub xy “
The defining equations are
÷
ø
ö
ç
è
æ
¶
¶
¶
¶
=
¶
¶
x
f
xx
f
2
2
, ÷
÷
ø
ö
ç
ç
è
æ
¶
¶
¶
¶
=
¶¶
¶
y
f
xyx
f
2
and so on. Notice the order in which the derivatives are taken:
yx
f
2
¶¶
¶
Differentiate first with respect to y, then with respect to x.
()
x
yyx
ff= Means the same thing.
Example:
1. Let ( )
6423
yxyxy,xf -= . Find
2
2
x
f
¶
¶
,
xy
f
2
¶¶
¶
,
2
2
y
f
¶
¶
and
yx
f
2
¶¶
¶
.
2. If ( )
x
yeycosxy,xf += , find
2
2
x
f
¶
¶
,
xy
f
2
¶¶
¶
,
2
2
y
f
¶
¶
and
yx
f
2
¶¶
¶
.
The Chain Rule
Chain Rule for Functions of Two Independent Variables
If ()y,xfw= has continuous partial derivatives xf and yf and if()txx=, ()tyy=
are
differentiable functions of t, then the compose ()()( )ty,txfw= is a differentiable
function of t and
()()( ) () ()()( ) (),tyty,txftxty,txf
dt
df
'
y
'
x
·+·=
or
dt
dy
y
f
dt
dx
x
f
dt
dw
¶
¶
+
¶
¶
= .
Example:
Use the chain rule to find the derivative of xyw=, with respect to t along the path

4. BMM 104: ENGINEERING MATHEMATICS I Page 4 of 8
tcosx= , tsiny= . What is the derivative’s value at
2
t
p
=?
Chain Rule for Functions of Three Independent Variables
If ( )z,y,xfw= is differentiable and x, y and z are 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
¶
¶
+
¶
¶
+
¶
¶
= .
Example:
Find
dt
dw
if zxyw += , tcosx= , tsiny= , tz=.
Chain Rule for Two Independent Variables and Three Intermediate Variables
Suppose that ( )z,y,xfw= , ()s,rgx= , ()s,rhy= , and ()s,rkz= . If all four
functions
are differentiable, then w has partial derivatives with respect to r and s, given by the
formulas
r
z
z
w
r
y
y
w
r
x
x
w
r
w
¶
¶
¶
¶
+
¶
¶
¶
¶
+
¶
¶
¶
¶
=
¶
¶
s
z
z
w
s
y
y
w
s
x
x
w
s
w
¶
¶
¶
¶
+
¶
¶
¶
¶
+
¶
¶
¶
¶
=
¶
¶
Example:
Express
r
w
¶
¶
and
s
w
¶
¶
in terms of r and s is
2
zy2xw ++= ,
s
r
x=, slnry
2
+= , r2z=.
If ()y,xfw= , ()s,rgx= , and ()s,rhy= , then
r
y
y
w
r
x
x
w
r
w
¶
¶
¶
¶
+
¶
¶
¶
¶
=
¶
¶
and
s
y
y
w
s
x
x
w
s
w
¶
¶
¶
¶
+
¶
¶
¶
¶
=
¶
¶
Example:
Express
r
w
¶
¶
and
s
w
¶
¶
in terms of r and s if

5. BMM 104: ENGINEERING MATHEMATICS I Page 5 of 8
22
yxw += , srx-= , sry+= .
If ()xfw= and ()s,rgx= , then
r
x
dx
dw
r
w
¶
¶
=
¶
¶
and
s
x
dx
dw
s
w
¶
¶
=
¶
¶
.
PROBLEM SET: CHAPTER 5
1. Sketch and name the surfaces
(a) ( )
222
zyxz,y,xf ++= (e) ( )
22
yxz,y,xf +=
(b) ( ) )zyxln(z,y,xf
222
++= (f) ( )
22
zyz,y,xf +=
(c) ( ) zxz,y,xf += (g) ( )
22
yxzz,y,xf --=
(d) ( )zz,y,xf = (h) ( )
9
z
16
y
25
x
z,y,xf
222
++=
2. Find
x
f
¶
¶
and
y
f
¶
¶
.
(a) ( ) 2y6x3yx7xy5y,xf
22
+-+--=
(b) ( ) ÷
ø
ö
ç
è
æ
=
-
x
y
tany,xf
1
(c) ( )
( )1yx
ey,xf
++
=
(d) ( ) ( )yxsiney,xf
x
+=
-
(e) ( ) ( )yxlny,xf +=
(f) ( ) ( )y3xsiny,xf
2
-=
3. Find x
f, y
f and
zf.
(a) ( ) ()xyzsinz,y,xf
1-
=
(b) ( )
( )
222
zyx
ez,y,xf
++-
=
(c) ( )
xyz
ez,y,xf
-
=
(d) ( ) ( )z3y2xtanhz,y,xf ++=
4. Find all the second-order partial derivatives of the following functions.
(a) ( ) xyyxy,xf ++=
(b) () xysiny,xf=
(c) ( ) xsinyycosyxy,xf
2
++=

6. BMM 104: ENGINEERING MATHEMATICS I Page 6 of 8
(d) ( ) 1yxey,xf
y
++=
5. Verify that yxxyww=.
(a) ( )y3x2lnw += (c)
43322
yxyxxyw ++=
(b) xlnyylnxew
x
++= (d) xyxsinyysinxw ++=
6. In the following questions, (a) express
dt
dw
as a function of t, both by using
the Chain Rule and by expressing w in terms of t and differentiating directly with
respect to t. The (b) evaluate
dt
dw
at the given value of t.
(i)
22
yxw += , tcosx= , tsiny= ; p=t.
(ii)
z
y
z
x
w+= , tcosx
2
= , tsiny
2
= ,
t
1
z= 3t=.
7. In the following questions, (a) express
u
z
¶
¶
and
v
z
¶
¶
as a functions of u and v
both by using the Chain Rule and by expressing z directly in terms of u and v
before differentiating. Then (b) evaluate
u
z
¶
¶
and
v
z
¶
¶
at the given point ()v,u.
(i) ylne4z
x
= , ( )vcosulnx= , vsinuy= ; ( ) ÷
ø
ö
ç
è
æ
=
4
,2v,u
p
(ii) ÷
÷
ø
ö
ç
ç
è
æ
=
-
y
x
tanz
1
, vcosux= , vsinuy= ; () ÷
ø
ö
ç
è
æ
=
6
,3.1v,u
p
ANSWERS FOR PROBLEM SET: CHAPTER 5
2. (a) ,3x14y5
x
f
+-=
¶
¶
6y2x5
y
f
--=
¶
¶
(b) ,
yx
y
x
f
22
+
-=
¶
¶
22
yx
x
y
f
+
=
¶
¶
(c)
( )
,e
x
f
1yx++
=
¶
¶

( )1yx
e
y
f
++
=
¶
¶
(d) ( ) ( ),yxcoseyxsine
x
f
xx
+++-=
¶
¶
--
( )yxcose
y
f
x
+=
¶
¶
-
(e) ,
yx
1
x
f
+
=
¶
¶

yx
1
y
f
+
=
¶
¶
(f) ( ) ( ),y3xcosy3xsin2
x
f
--=
¶
¶
( ) ( )y3xcosy3xsin6
x
f
---=
¶
¶

7. BMM 104: ENGINEERING MATHEMATICS I Page 7 of 8
3. (a)
222
x
zyx1
yz
f
-
=
,
222
y
zyx1
xz
f
-
=
,
222
z
zyx1
xy
f
-
=

(b)
( )
222
zyx
x xe2f
++-
-= ,
( )
222
zyx
y ye2f
++-
-= ,
( )
222
zyx
z
ze2f
++-
-=
(c)
xyz
x
yzef
-
-= ,
xyz
y
xzef
-
-= ,
xyz
zxyef
-
-=
(d) ( )z3y2xhsecf
2
x ++= , ( )z3y2xhsec2f
2
y ++= ,
( )z3y2xhsec3f
2
z ++=
4. (a) ,y1
x
f
+=
¶
¶
x1
y
f
+=
¶
¶
, ,0
x
f
2
2
=
¶
¶
,0
y
f
2
2
=
¶
¶
1
yx
f
xy
f
22
=
¶¶
¶
=
¶¶
¶
(b) ,xycosy
x
f
=
¶
¶
xycosx
y
f
=
¶
¶
, ,xysiny
x
f
2
2
2
-=
¶
¶
,xysinx
y
f
2
2
2
-=
¶
¶
xysinxyxycos
yx
f
xy
f
22
-=
¶¶
¶
=
¶¶
¶
(c) ,xcosyxy2
x
f
+=
¶
¶
xsinysinx
y
f
2
+-=
¶
¶
, ,xsinyy2
x
f
2
2
-=
¶
¶

,ycos
y
f
2
2
-=
¶
¶
xcosx2
yx
f
xy
f
22
+=
¶¶
¶
=
¶¶
¶
(d)
y
e
x
f
=
¶
¶
1xe
y
f
y
+=
¶
¶
, ,0
x
f
2
2
=
¶
¶
,xe
y
f
y
2
2
=
¶
¶
y
22
e
yx
f
xy
f
=
¶¶
¶
=
¶¶
¶
5. (a) ,
y3x2
2
x
w
+
=
¶
¶
,
y3x2
3
y
w
+
=
¶
¶
( )
,
y3x2
6
xy
w
2
2
+
-
=
¶¶
¶
and
( )
2
2
y3x2
6
yx
w
+
-
=
¶¶
¶

(b) ,
x
y
ylne
x
w
x
++=
¶
¶
,xln
y
x
y
w
+=
¶
¶
,
x
1
y
1
xy
w
2
+=
¶¶
¶
and
x
1
y
1
yx
w
2
+=
¶¶
¶


(c) ,yx3xy2y
x
w
4232
++=
¶
¶
,yx4yx3xy2
y
w
3322
++=
¶
¶
,yx12xy6y2
xy
w
322
2
++=
¶¶
¶
and
322
2
yx12xy6y2
yx
w
++=
¶¶
¶

8. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
(d) ,yxcosyysin
x
w
++=
¶
¶
,xxsinycosx
y
w
++=
¶
¶
,1xcosycos
xy
w
2
++=
¶¶
¶
and 1xcosycos
yx
w
2
++=
¶¶
¶


6. (i)(a) 0
dt
dw
= (b)0
(ii)(a) 1
dt
dw
= (b)1

7. (i)(a) ( )( ) vcos4vsinulnvcos4
u
z
+=
¶
¶
( )( )
vsin
vcosu4
vsinulnvsinu4
v
z
2
+-=
¶
¶
(b) ( )22ln2
u
z
+=
¶
¶
242ln22
v
z
+-=
¶
¶
(ii)(a) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶
(b) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶

9. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
(d) ,yxcosyysin
x
w
++=
¶
¶
,xxsinycosx
y
w
++=
¶
¶
,1xcosycos
xy
w
2
++=
¶¶
¶
and 1xcosycos
yx
w
2
++=
¶¶
¶


6. (i)(a) 0
dt
dw
= (b)0
(ii)(a) 1
dt
dw
= (b)1

7. (i)(a) ( )( ) vcos4vsinulnvcos4
u
z
+=
¶
¶
( )( )
vsin
vcosu4
vsinulnvsinu4
v
z
2
+-=
¶
¶
(b) ( )22ln2
u
z
+=
¶
¶
242ln22
v
z
+-=
¶
¶
(ii)(a) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶
(b) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶

10. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
(d) ,yxcosyysin
x
w
++=
¶
¶
,xxsinycosx
y
w
++=
¶
¶
,1xcosycos
xy
w
2
++=
¶¶
¶
and 1xcosycos
yx
w
2
++=
¶¶
¶


6. (i)(a) 0
dt
dw
= (b)0
(ii)(a) 1
dt
dw
= (b)1

7. (i)(a) ( )( ) vcos4vsinulnvcos4
u
z
+=
¶
¶
( )( )
vsin
vcosu4
vsinulnvsinu4
v
z
2
+-=
¶
¶
(b) ( )22ln2
u
z
+=
¶
¶
242ln22
v
z
+-=
¶
¶
(ii)(a) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶
(b) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶

11. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
(d) ,yxcosyysin
x
w
++=
¶
¶
,xxsinycosx
y
w
++=
¶
¶
,1xcosycos
xy
w
2
++=
¶¶
¶
and 1xcosycos
yx
w
2
++=
¶¶
¶


6. (i)(a) 0
dt
dw
= (b)0
(ii)(a) 1
dt
dw
= (b)1

7. (i)(a) ( )( ) vcos4vsinulnvcos4
u
z
+=
¶
¶
( )( )
vsin
vcosu4
vsinulnvsinu4
v
z
2
+-=
¶
¶
(b) ( )22ln2
u
z
+=
¶
¶
242ln22
v
z
+-=
¶
¶
(ii)(a) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶
(b) 0
u
z
=
¶
¶
1
v
z
-=
¶
¶