Solution Exercise 5 Session 4

Exercise Session 4, Solution, November 3th , 2006 Mathematics for Economics and Finance Prof: Norman Schürho¤ TAs: Zhihua (Cissy) Chen, Natalia Guseva Exercise 5 Let f : R2 ! R be ( p xy x2 +y 2 f (x; y) = 0

if (x; y) 6= (0; 0); if (x; y) = (0; 0):

a) Is f a continuous function? Show. b) Is f a di¤erentiable function? Show. Solution (a) In all points but for (0; 0) function f is undoubtfully continuous. Let’s check at (0; 0) : lim

(x;y)!(0;0)

f (x; y) =

hence f (0; 0) =

lim

(x;y)!(0;0)

lim

(x;y)!(0;0)

p xy 2

x +y 2

=

lim

(x;y)!(0;0)

q

1 + x12

1 y2

= 0;

f (x; y): So, function is continuous.

(b) check whether partial derivatives exists: 3 when (x; y) 6= (0; 0); fx0 (x; y) = 2 y 2 3 ; fy00 (x; y) = (x +y ) 2

x3 3

(x2 +y 2 ) 2

when (x; y) = (0; 0); fx0 (0; 0) = lim f (x;0) x f (0;0) = 0; fy0 (0; 0) = lim f (0;y) y f (0;0) = y!0

x!0

0 =) all partial derivatives exist. In all points but for (0; 0) function f is undoubtfully di¤erentiable. If f is di¤erentiable in (0; 0), when the function has a derivative in the direction of any unit vector and Df(x;y) (0; 0) = fx0 (0; 0)x + fy0 (0; 0)y where p x2 + y 2 = 1 kf (x;y) f (0;0)k Take direction x = y, we have Dfx=y (0; 0) = lim = k(x;y) (0;0)k (x;y)!(0;0)

lim

p

xx x2 +x2

x fx0 (0; 0)

x!0

0

2 = lim px2x2 x!0 = fy0 (0; 0) =

=

p1 2

0 =) fx0 (0; 0)x + fy0 (0; 0)y = 0 hence function can not be di¤erentiable in (0; 0):

1

p1 2

6= 0 and