Functions of Several Variables

Dr. Alpesh M. Dhorajia

Birla Institute of Technology and Science,

K K Birla Goa Campus.

Partial Derivatives

First Partial Derivatives

First Partial Derivatives of f(x, y)

Suppose f(x, y) is a function of two variables x and y.

Then, the first partial derivative of f with respect to xat the point (x, y) is

provided the limit exists.

The first partial derivative of f with respect to y at the point (x, y) is

provided the limit exists.

0

( , ) ( , )limh

f f x h y f x y

x h

0

( , ) ( , )limk

f f x y k f x y

y k

Second Order Partial Derivatives

The first partial derivatives fx(x, y) and fy(x, y) of a function

f(x, y) of two variables x and y are also functions of x and y.

As such, we may differentiate each of the functions fx and

fy to obtain the second-order partial derivatives of f.

Second Order Partial Derivatives

Differentiating the function fx with respect to x leads to the

second partial derivative

But the function fx can also be differentiated with respect

to y leading to a different second partial derivative

2

2( )xx x

ff f

x x

2

( )xy x

ff f

y x y

Second Order Partial Derivatives

Similarly, differentiating the function fy with respect to y

leads to the second partial derivative

Finally, the function fy can also be differentiated with

respect to x leading to the second partial derivative

2

2( )yy y

ff f

y y

2

( )yx y

ff f

x y x

Second Order Partial Derivatives

Thus, four second-order partial derivatives can be obtained of a function of two variables:

f

x

y

f

x

f

y

x

y

2f f

x y x y

2

2

f f

y y y

x

y

2

2

f f

x x x

2f f

y x y x

2 2f f

y x x y

When both are continuous

Examples

Find the second-order partial derivatives of the function

Solution

First, calculate fx and use it to find fxx and fxy:

3 2 2 2( , ) 3 3f x y x x y xy y

2 23 6 3x xy y

3 2 2 2( 3 3 )xf x x y xy yx

2 2(3 6 3 )xxf x xy yx

6 6x y

2 2(3 6 3 )xyf x xy yy

6 6x y

6( )x y 6( )y x

Examples

Find the second-order partial derivatives of the function

Solution

Then, calculate fy and use it to find fyx and fyy:

3 2 2 2( , ) 3 3f x y x x y xy y

23 6 2x xy y

3 2 2 2( 3 3 )yf x x y xy yy

2( 3 6 2 )yxf x xy yx

6 6x y

2( 3 6 2 )yyf x xy yy

6 2x

6( )y x 2(3 1)x

Let

find and at (0,0). Check whether ?

How to find ? Can we find directly? (why not?)

We have to use the definition:

At (x,y) = (0,0)

2 2

2 2, ( , ) (0,0)

( , )

0, ( , ) (0,0)

x yxy x y

f x y x y

x y

xyf yxfxy yxf f

xf

0 0

( , ) ( , ) (0,0 ) (0,0)lim limx x x x

xyk k

f x y k f x y f k ff

k k

0 0

( ,0) (0,0) 0 0(0,0) lim lim 0x

h h

f h ff

h h

2 2

2 2

0 0

( , ) (0, )(0, ) lim limx

h h

h khk

f h k f k h kf k kh h

We have at (0,0)

Similarly at (0,0) is = 1.

Therefore at (0,0) at (0,0).

Question: Are the both and are continuous?

Verify using directly computing !!!

0 0

(0, ) (0,0) 0lim lim 1x x

xyk k

f k f kf

k k

yxf

xy yxf f

xyf yxf

Consider

Find and at (0,0) if it exist.

Does this limit exists? (Ans: No.) Why?

Similarly, at (0,0) ……………. ?

2sin ( ), ( , ) (0,0)

( , ) | | | |

0, ( , ) (0,0)

x yx y

f x y x y

x y

xfyf

fx(0,0) = lim

h®o

f (0+ h,0) - f (0,0)

h

= limh®0

sin2 h

| h |-0

h= limh®0

sin2 h

h | h |

yf

Show that none of the partial derivatives exist at (0,0) although the function is continuous at (0,0).

sin(1/ ) sin(1/ ), 0, 0,

sin(1/ ), 0, 0( , )

sin(1/ ), 0, 0

0, ( , ) (0,0)

x x y y x y

x x x yf x y

y y y x

x y

By definition

Similarly,

0 0

0

1sin 0

(0 ,0) (0,0)(0,0) lim lim

1limsin ................?

xh h

h

hf h f hf

h h

h

0 0

0 0

(0,0 ) (0,0) (0, ) (0,0)(0,0) lim lim

1sin 0

1lim limsin ..............?

yk k

k k

f k f f k ff

k k

kk

k k

Continuity of function at (0,0).

If

Take .

Similarly: if and limit exist.

2 2

1 1| ( , ) (0,0) | | sin sin 0 |

1 1| sin | | sin | | | | | 2

f x y f x yx y

x y x y x yx y

0, 0,( , ) (0,0).x y x y

2

0, 0x y 0, 0x y

Give an example of a function in two variables such

that

(1) Function is continuous but partial derivatives does not exist.

(2) Both partial derivatives are exist but the function is not continuous.

(3) Mixed partial derivatives are not equal.

Differentiability of Functions of two variables

Recall the definition of derivative in one variable case:

i.e.

Definition: Let be a function of two variables and is said to be differentiable at (x,y) if there exist a pair such that

'

0

( ) ( )lim ( )h

f x h f xf x

h

'

0

( ) ( ) ( )lim 0h

f x h f x hf x

h

: Rf D

lim(h,k )®(0,0)

f (x+h, y+ k)- f (x, y)-a1h-a

2k

h2 + k 2= 0

(a1,a

2)

What is if f is a differentiable function?

Since f is differentiable

If we choose a path (h,0) tending to (0,0), then

i.e. and similarly .

If f is differentiable then is called the total derivative of f(x,y).

1 2( , )

1 2

2 2( , ) (0,0)

( , ) ( , )lim 0

h k

f x h y k f x y h k

h k

1 2

2 2( , ) (0,0)

( , ) ( , ) 0lim 0

0h k

f x h y f x y h

h

1

f

x

2

f

y

1 2df dx dy

To check given function is differentiable at (x , y)

(1) and should exist.

(2) .

f

x

f

y

2 2( , ) (0,0)

( , ) ( , )lim 0

x y

h k

f x h y k f x y hf kf

h k

Consider to show f is differentiable at every points (x , y).

Clearly and are exist.

To show .

2 2( , )f x y x y

2f

xx

2

fy

y

2 2( , ) (0,0)

( , ) ( , )lim 0

x y

h k

f x h y k f x y hf kf

h k

2 2 2 2

2 2( , ) (0,0)

2 2 2 2 2 2

2 2( , ) (0,0)

2 22 2

2 2( , ) (0,0) ( , ) (0,0)

( ) ( ) (2 ) (2 )lim

2 2 (2 ) (2 )lim

lim lim 0

h k

h k

h k h k

x h y k x y h x k k

h k

x xh h y yk k x y h x k k

h k

h kh k

h k