Differentiability of f(x,y) = (x^2 y)/(x^4+y^2) if (x,y) not

[wickeddude12]

Apparently, the function is not differentiable at (0,0). However, from my calculations the first-order partial derivatives exist and are continuous at (0,0) (which imply differentiability). Would someone show that the initial requirement of the theorem does not hold?

 

[galactus]

Re: Differentiability

We can show the limit does not exist by letting $\displaystyle (x,y)\rightarrow (0,0)$ along the parabola $\displaystyle y=x^{2}$

We get $\displaystyle {\displaystyle}\lim_{x\to 0}\frac{x^{4}}{2x^{4}}=\lim_{x\to 0}\frac{1}{2}=\frac{1}{2}$

Also, if we let $\displaystyle (x,y)\rightarrow (0,0)$ along any line y=mx, then

$\displaystyle \lim_{x\to 0}\frac{mx^{3}}{x^{4}+m^{2}x^{2}}=\lim_{x\to 0}\frac{mx}{x^{2}+m^{2}}=0$

The limit does not exist.

 

[wickeddude12]

Re: Differentiability

Thank you, but you did not answer my question. Would you be able to show me that the first-order partial derivatives are not continuous (or do not exist) at (0,0)?

EDIT: Question solved: I just proved to myself that the first order partial derivative of f with respect to x is discontinuous at (0,0) which DOES NOT imply that f is not differentiable there.

 

[galactus]

Re: Differentiability

I'm sorry, but that is kind of what I was eluding to.

To prove the partials exist but f is not continuous at (0,0), we can show that $\displaystyle f_{x}(0,0)=0, \;\ f_{y}(0,0)=0$ by expressing the derivatives as limits.

Remember, $\displaystyle f_{x}(x,y)={\displaystyle}\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$

and

$\displaystyle f_{y}(x,y)=\lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}$

To prove f is not continuous at (0,0), show that $\displaystyle \lim_{(x,y)\to (0,0)}f(x,y)$ does not exist by letting $\displaystyle (x,y)\rightarrow (0,0)$ along y=0 and along y=x^2.

Along y=0, $\displaystyle \lim_{x\to 0}\frac{0}{x^{2}}=0$

Alomg y=x^2, $\displaystyle \lim_{x\to 0}\frac{x^{4}}{x^{4}+x^{4}}=\frac{1}{2}$

I hope this helps. Is this what you meant?.

 

[wickeddude12]

Re: Differentiability

I just realized that we cannot show f is not differentiable at (0,0) by proving that a partial derivative is discontinuous there, but we can prove f is differentiable by proving that the partial derivative is continuous at (0,0). Therefore the only way is to show that the function itself is discontinuous at (0,0). Therefore, my original question was useless.