DIFFERENTIABILITY OF REAL VALUED FUNCTIONS OF TWO VARIABLES AND EULER’S THEOREM ARUN LEKHA Associate Professor G.C.G., SECTOR-11, CHANDIGARH
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DIFFERENTIABILITY OF REAL VALUED FUNCTIONS OF TWO VARIABLES AND
EULER’S THEOREM
ARUN LEKHA
Associate Professor
G.C.G., SECTOR-11, CHANDIGARH
FUNCTION OF TWO VARIABLES
Definition: A variable Z is said to be a function of two independent variables x and y denoted by z=f (x,y) if to each pair of values of x and y over some domain Df ={(x,y): a<x<b,c<y<d} there corresponds a single definite value of Z.
e.g. The area A of a rectangle having sides of lengths x and y is xy.
i.e. A = xy is a function of two variables x and y. Domain of the function is Df={(x,y): x > 0 y > 0}.
PARTIAL DERIVATIVE (FIRST ORDER)
Partial derivative of Z = f(x,y) w.r.t.x, regarding y as constant is denoted by
z/x or f/x or fx and
provided it exists and is finite.
Similarly
provided it exists and is finite.
x
yxfyxxf
x
ltf x
),(),(
0
y
yxfyyxf
y
ltf y
),(),(
0
e.g. If z =e-x/y + tan-1 (x/y).
then
PARTIAL DERIVATIVES (SECOND ORDER)
The first order partial derivatives z/x or z/ y are generally functions of x and y and hence we can again find their partial derivatives w.r.t. X or y. The partial derivatives thus obtained are called second order partial derivatives and are denoted by fxy or fyx.
22
/
yx
y
y
e
x
z yx
HOMOGENEOUS FUNCTIONS
A function of two variables x and y of the form
f(x,y) = aoxn+a1x
n-1 y + ….an-1 xyn-1+anyn
in which each term is of degree n is called homogeneous function or if it can be expressed in the form yng(x/y) or xng(y/x).
e.g. f(x,y) = x2+y2 / x+y
is homogeneous function of degree 1
EULER’S THEOREM: If Z = f(x,y) is a homogeneous function of x and y of degree n , then
x z/x + y z/ y = nz
Ex. Show that Z =ax2 + 2hxy + by2 is homogeneous function of degree 2 and verify Euler’s theorem.
Sol. Z=ax2+2hxy + by2 = x2 [a+2h y/x + by2/x2] = X2 g (y/x).
Z is homo. Function of degree 2.
Verification of Euler’s theorem.
z/ x = 2ax + 2hy.
z/ y = 2hx + 2by
then x z/x + yz/y
= x(2ax+2hy) + y (2hx+2by)
=2ax2 + 4 hxy + 2by2
= 2z
Euler’s theorem for a homogeneous function of three independent variables.
If H is a homogeneous function or x,y,z of order n then x H/x + Y h/Y + z H/Z = nH
Differentiable Function: A function of (x,y) is said to be differentiable at (x,y) if z = f (x,y) can be expressed in the form
z = /x f(x,y). x + /yf (x,y). y
+ 1 y + 2 x
where 1 0, 2 0 as x, y 0.
Remark: Continuity of f, fx, fy at (x,y) are
sufficient conditions for differentiability. Total differential of a function If z = f (x,y), then total differential of Z is denoted
and defined by dz = z/x. dx + z/y.dy
Differentiability of f (x,y)
The function Z=f (x,y) is said to be differentiable at a point (xo,yo) if in a neighbourhood of (xo,yo), it can be represented in the form
f (xo+h, yo+k) – f (xo,yo) = Ah + BK + h +k
were A,B are independent of the variables h,k.
and , 0 as h,k 0 independently.
Thm: If a function f(x,y) is differentiable at a point (xo,yo), then it is continuous at that point.
Remark: Converse is not always true.
Example: f (x,y) = |xy| is not differentiable at (0,0) but continuous at (0,0).
Theorem: If a function f (x,y) is differentiable at a point (xo,yo) then fx (xo,yo) and fy (xo,yo) both exist and
Ah
yxfyhxf
h
Ltyxf oooo
oox
),(),(
0),(
Bk
yxfkyx
k
Ltyxf oooo
ooy
),(),(
0),(
Que. Discuss the differentiability of
f (x1y) = |x|+|y| at (0,0)
Sol. f(x,y) is differentiable at (0,0) If f (0+h,0+k) – f(0,0) = Ah+Bk +h +k where ,0 as h, k0
01
0,1||
0
|)0||0(||0|||
0
)0,0()0,(
0)0,0(
hif
hif
h
h
h
Lt
h
h
h
Lt
h
fhf
h
LtfANow x
A does not exist.
Similarly B =fy (0,0) =
B does not exist.
Hence f(x,y) is not differentiable at (0,0).
YOUNG’S THEOREM
Let f(x) be defined in a domain D R2. Let (a,b) be an interior of D and let
(i) fx and fy exist in the neighbourhood of (a,b)
(ii) fx and fy are differentiable at the point (a,b)
then fxy = fyx at (a,b)
01
0,1||
0
)0,0(),0(
0
k
k
k
k
k
Lt
k
fkf
k
Lt
SCHAWARZ’S THEOREM If (a,b) be a point of the domain DR2 of
a function f (x,y) such that
(i) fx and fy exist in the neighbourhood of the point (a,b)
(ii) fxy is continuous at (a,b)
then fyx exist at (a,b)
and fyx =fxy at (a,b)
Change of Variables
Let Z = f(x,y), x = (u,v), y = (u,v)
Taking as constant,
Taking u as constant
and by solving the above equations in
z/x, z/y we get their values in terms of
z/u, z/v, u, v.
u
y
y
z
u
x
x
z
u
z
..
v
y
y
z
v
x
x
z
v
z
..
Composite functions:
Definition: Let Z = f (x,y) and let x = (t)
and y = (t), then z is called composite function of t.
Differentiation of composite functions:
Let Z = f (x,y) possess continuous partial derivatives and
X = (t), y = (t) possess continuous derivatives,
then dz/dt = z/x. dx/dt + z/y. dy/dt
Implicit functions:
Definition: Let f(x,y) be a function of two variables and
y = (x) be a function of x such that f (x,(x)) vanishes identically, then y = (x) is an implicit function defined by the functional equation
f(x,y) = 0
Differentiation of implicit functions
If f(x,y) = 0 or c be an implicit function then
(i) dy/dx = -f/x / f/y = -fx/fy, fy 0
(ii) d2y/dx2 =
Implicit function theorem (Two Variables)
Let f(x,y) be a function of two variables
x and y and (a,b) be a point of its domain of definition such that
(i) f(a,b) = 0
(ii) fx and fy exist and are continuous in certain nbhd. of (a,b).
0,)(
)(2)(3
22
fy
fy
ffffffyfxx xyyyxyx
(ii) fy (a,b) 0, then there exist a rectangle (a-h, a+h, b-k, b+k) about (a,b) such that for every x in the interval [a-h, a+h], f (x,y) = 0 determines one and only one value
y = (x) lying in the internal [b-k, b+k] with the following properties
(i) b = (a)
(ii) f(x, (x)) = 0 for every x in [a-h,a+h]
(iii) (x) is derivable and both (x) and ’(x) are continuous in [a-h, a+h].
e.g. f(x,y) = x2+y2-1 and a point (0,1)
So that f(0,1) = 0 and fy (0,1) = 2 0
Now of the two possible solutions
y = + 1 – x2
(i) y = + 1-x2 is implicit function in nbhd. of (0,1), where |x|<1, y>0.
(ii) y = - 1-x2 is implicit function in nbhd. of (0,-1) where |x|<1 , y<0.
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