# Show the the derivative exists but not continuous

• Aug 11th 2010, 03:42 AM
tsang
Show the the derivative exists but not continuous
Hi dear friends here, I got a question, it supposed to be quite simple, but it is a little bit tricky, can anyone please help me? Thanks a lot.
Here it is:

f(x,y)={x^3/(x^6+y^2), for (x,y) does not equal (0,0)}
f(x,y)={0, for (x,y)=(o,o)}
Show that df/dx exists at (0,0) but is not continuous there.

I used first principle to prove the limit of $df/dx$is 0 at (x,y)=(0,0), which means it does exist, am I right? What is the exact definition to say something exists?
Then, I can imagine it is discontinuous because at (0,0) the graph of $df/dx$should be broken, which has the circle there, but I just forgot how to show it. I thought the definition of "continuiy" is the value of limit equals to the value of function. So how do I write about it? Please help me, thanks a lot.
• Aug 11th 2010, 03:56 AM
HallsofIvy
Quote:

Originally Posted by tsang
Hi dear friends here, I got a question, it supposed to be quite simple, but it is a little bit tricky, can anyone please help me? Thanks a lot.
Here it is:

f(x,y)={x^3/(x^6+y^2), for (x,y) does not equal (0,0)}
f(x,y)={0, for (x,y)=(o,o)}
Show that df/dx exists at (0,0) but is not continuous there.

I used first principle to prove the limit of $df/dx$is 0 at (x,y)=(0,0), which means it does exist, am I right? What is the exact definition to say something exists?

You mean $\frac{\partial f}{\partial y}$ here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

Yes, thi partial derivative exists at x= 0 if and only if $\lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h}$ exists.

In this case, that limit is $\lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7}$ $= \lim_{h\to 0}\frac{1}{h^4}$. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?

Quote:

Then, I can imagine it is discontinuous because at (0,0) the graph of $df/dx$should be broken, which has the circle there, but I just forgot how to show it. I thought the definition of "continuiy" is the value of limit equals to the value of function. So how do I write about it? Please help me, thanks a lot.
• Aug 11th 2010, 04:06 AM
tsang
Quote:

Originally Posted by HallsofIvy
You mean $\frac{\partial f}{\partial y}$ here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

Yes, thi partial derivative exists at x= 0 if and only if $\lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h}$ exists.

In this case, that limit is $\lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7}$ $= \lim_{h\to 0}\frac{1}{h^4}$. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?

Thank you. I'm sure I copied down the right question. Also, I do mean by $\frac{\partial f}{\partial y}$, sorry that I'm not very good at with using code.
I think the question is trying to say $\frac{\partial f}{\partial y}$ is exists but discontinues, not original function. So what I did was differentiate original function first, and I end up with [tex]\frac{\partial f}{\partial y}=(3x^2y^3-3x^8y)/((x^6+y^2)^2)[tex], and use first principle as you did, so I can find the limit equal to 0.
But not sure how to sure discontinues.
• Aug 11th 2010, 04:12 AM
tsang
Quote:

Originally Posted by HallsofIvy
You mean $\frac{\partial f}{\partial y}$ here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

Yes, thi partial derivative exists at x= 0 if and only if $\lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h}$ exists.

In this case, that limit is $\lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7}$ $= \lim_{h\to 0}\frac{1}{h^4}$. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?

thank you, I'm sure I got right question, and it is $\frac{\partial f}{\partial y}$, sorry that I'm not very good with using code.
I think the question is asking about $\frac{\partial f}{\partial y}$, rather than original function. I differentiate original function first, and I end up with $df/dx=(3x^2y^3-3x^8y)/(x^6+y^2)^2$, which can find limit at (0,0) is 0.
But I'm not sure how to show it is discontinues.