Let f:R^2->R be defined as f(x,y)={(x^2+y^2)cos(1/(sqrt(x^2+y^2)), for (x,y) not = (0,0), 0, forfor (x,y) not = (0,0) then check whether its differentiable and also whether its partial derivatives ie f_x,f_y are continuous at (0,0).

Laila Murphy

Laila Murphy

Answered question

2022-11-17

Let f : R 2 R be defined as
f ( x , y ) = { ( x 2 + y 2 ) cos 1 x 2 + y 2 , for  ( x , y ) ( 0 , 0 ) 0 , for  ( x , y ) = ( 0 , 0 )
then check whether its differentiable and also whether its partial derivatives ie f x , f y are continuous at ( 0 , 0 ). I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for f x as function is symmetric in y and x. So f x turns out to be
f x ( x , y ) = 2 x cos ( 1 x 2 + y 2 ) + x x 2 + y 2 sin ( 1 x 2 + y 2 )
which is definitely not 0 as ( x , y ) ( 0 , 0 ). Same can be stated for f y . But how to proceed with the first part?

Answer & Explanation

kavdawg8w8

kavdawg8w8

Beginner2022-11-18Added 20 answers

The partial derivatives at ( 0 , 0 ) are x f ( 0 , 0 ) = 0 and y f ( 0 , 0 ) = 0.
The only candidate for the differential at ( 0 , 0 ) is:
D f ( 0 , 0 ) = [ x f ( 0 , 0 ) y f ( 0 , 0 ) ] = [ 0 0 ]
which is simply the zero-functional.
We have:
lim ( h 1 , h 2 ) ( 0 , 0 ) | f ( h 1 , h 2 ) f ( 0 , 0 ) D f ( 0 , 0 ) ( h 1 , h 2 ) | ( h 1 , h 2 ) = lim ( h 1 , h 2 ) ( 0 , 0 ) h 1 2 + h 2 2 h 1 2 + h 2 2 | cos 1 h 1 2 + h 2 2 | = lim ( h 1 , h 2 ) ( 0 , 0 ) h 1 2 + h 2 2 | cos 1 h 1 2 + h 2 2 | 1 = 0
Hence, f is differetiable at ( 0 , 0 ) with the differential being 0, even though the partial derivatives are not continuous at ( 0 , 0 ).
reevelingw97

reevelingw97

Beginner2022-11-19Added 4 answers

we have
f ( 0 , 0 ) y = lim h 0 h cos ( 1 h ) = 0
since | h cos ( 1 h ) | | h | which tends to 0

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