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mixed partial derivatives

Unparseable latex formula:

f(x,y) = [br] \begin{array}{l l}[br] \dfrac{xy^3}{x^2 + y^2} & \quad (x,y) \neq (0,0)\\[br] 0 & \quad (x,y) = (0,0)[br] \end{array} \right.



I need to show that fxy(0,0)=1f_{xy}(0,0) = 1.

I did
Unparseable latex formula:

f_{xy}(x,y) = [br] \begin{array}{l l}[br] \dfrac{-3x^4y^2 + 6x^2y^4 + y^6}{(x^2 + y^2)^3} & \quad (x,y) \neq (0,0)\\[br] 0 & \quad (x,y) = (0,0)[br] \end{array} \right.




Clearly fxy(0,0)=0f_{xy}(0,0) = 0 . What am i doing wrong?
(edited 11 years ago)
Reply 1
Avatar for DFranklin
DFranklin
18
Well, how did you work out that fxy(0,0)=0f_{xy}(0, 0) = 0? You seem to have just written it down.

For comparison:

Let g(x) = x (x non-zero), g(x) = 0 (x=0).

Obviously g'(0) is NOT found by "differentiating" g(x) = 0
Reply 2
Avatar for amy1e3
amy1e3
OP
Original post by DFranklin
Well, how did you work out that fxy(0,0)=0f_{xy}(0, 0) = 0? You seem to have just written it down.

For comparison:

Let g(x) = x (x non-zero), g(x) = 0 (x=0).

Obviously g'(0) is NOT found by "differentiating" g(x) = 0


So you can't write
Unparseable latex formula:

f_{x}(x,y) = [br] \begin{array}{l l}[br] \dfrac{y^3(y^2 -x^2)}{(x^2 + y^2)^2} & \quad (x,y) \neq (0,0)\\[br] 0 & \quad (x,y) = (0,0)[br] \end{array} \right.

Reply 3
Avatar for DFranklin
DFranklin
18
Again, where did that come from? (in particular, where did the value for (x, y) = 0 come from?). If you don't explain what you're doing, how do you expect people to help?
Reply 4
Avatar for amy1e3
amy1e3
OP
Original post by DFranklin
Again, where did that come from? (in particular, where did the value for (x, y) = 0 come from?). If you don't explain what you're doing, how do you expect people to help?



Basically i differentiated the two parts separately, and of course the differential of 0 with respect to x is just 0.
Reply 5
Avatar for james22
james22
16
Original post by amy1e3
Basically i differentiated the two parts separately, and of course the differential of 0 with respect to x is just 0.


You cannot do this, I think you have to sonsider limits but I'm not to sure on this myself.
Original post by amy1e3
I need to show that fxy(0,0)=yf_{xy}(0,0) = y


Show that fx(0,y)=yf_x (0,y)=y then evaluate the second derivative from definition.
Reply 7
Avatar for ztibor
ztibor
10
Original post by amy1e3
Unparseable latex formula:

f(x,y) = [br] \begin{array}{l l}[br] \dfrac{xy^3}{x^2 + y^2} & \quad (x,y) \neq (0,0)\\[br] 0 & \quad (x,y) = (0,0)[br] \end{array} \right.



I need to show that fxy(0,0)=1f_{xy}(0,0) = 1.

I did
Unparseable latex formula:

f_{xy}(x,y) = [br] \begin{array}{l l}[br] \dfrac{-3x^4y^2 + 6x^2y^4 + y^6}{(x^2 + y^2)^3} & \quad (x,y) \neq (0,0)\\[br] 0 & \quad (x,y) = (0,0)[br] \end{array} \right.


Clearly fxy(0,0)=0f_{xy}(0,0) = 0 . What am i doing wrong?


1. WHat is fx′(0,0)f'_x(0,0)
2. What is fx′(x,y)f'_x(x,y)
2. What is f"xy(0,0)f"_{xy}(0,0)
For 1.
fx′(0,0)=limh→0f(h,0)−f(0,0)h=limh→00−0h=0f'_x(0,0)=lim_{h \rightarrow 0}\frac{f(h,0)-f(0,0)}{h}=lim_{h \rightarrow 0}\frac{0-0}{h}=0
so
2.
Unparseable latex formula:

f'_x(x,y)=\begin {Bmatrix} \frac{y^5-x^2y^3}{(x^2+y^2)^2} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end



It is continouos at (0,0)

3.
f"xy(0,0)=limk→0fx′(0,k)−fx′(0,0)k=limk→0k5k4=0≠1f"_{xy}(0,0)=lim_{k \rightarrow 0}\frac{f'_x(0,k)-f'_x(0,0)}{k}=lim_{k \rightarrow 0}\frac{k^5}{k^4}=0 \neq 1

Do you wrote all the question correctly?
(edited 11 years ago)
Original post by ztibor

fxy(0,0)=lim⁡k→0fx(0,k)−fx(0,0)k=lim⁡k→0k5k4=0≠1f_{xy}(0,0)=\displaystyle\lim_{k \rightarrow 0}\frac{f_x(0,k)-f_x(0,0)}{k}=\boxed{\lim_{k \rightarrow 0}\frac{k^5}{k^4}}=0 \neq 1


Mistake there.

Edit: also, you've essentially done all the work for her, which is not how this forum works.
(edited 11 years ago)
Reply 9
Avatar for ztibor
ztibor
10
Original post by Lord of the Flies
Mistake there.

Edit: also, you've essentially done all the work for her, which is not how this forum works.


Sorry and thanks

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