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.