# Evaluating $f_x(0,0)$ and $f_y(0,0)$ using the limit definition of the partial derivative [duplicate]

Possible Duplicate:
Evaluate the partial derivatives of the following function:

The function $f:\mathbb R^2 \to \mathbb R$, defined as: \left\{\begin{align*}&\frac{x^5y}{x^4+y^2}&&(x,y) \neq 0\\&0&&(x,y)=(0,0)\end{align*}\right. Using the limit definition I get:

$$f_x(0,0) = \frac{(0+h)^2 (0)}{h(0+h)^4+(0)} - \frac{(0)^5(0)}{h(0)^4+h(0)^2}$$

for both $f_x$ and $f_y$, I get $\frac00$ terms for the last term. I know that the answer to both is $0$, but how do I deal with the $\frac00$ term (I can't seem to get rid of it). Is this allowed?

Thanks for any insight!

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## marked as duplicate by Pedro Tamaroff♦, LVK, William, Thomas, Noah SnyderOct 4 '12 at 22:52

Please check if I haven't changed your question unintentionally. – Gigili Jun 6 '12 at 11:10
Are you aware of L'Hôpital's rule? – user22805 Jun 6 '12 at 11:44
@DavidWallace I guess that our friend should write down the correct expression for the derivative. It will be clear that the computation is elementary. – Siminore Jun 6 '12 at 11:48
@PeterTamaroff I just wanted to clarify something else to do with the question. Didn't know if I would get any help from amending the first post, so I posted again, specifically with where I needed help. – JackReacher Jun 7 '12 at 2:18
@DavidWallace - I was attempting to answer the question using the limit definition of the partial derivative, I don't think L'Hopital's rule is applicable to this question. – JackReacher Jun 7 '12 at 2:20

For $h\ne 0$ you should have
$$\frac{f(0+h,0)-f(0,0)}h=\frac{f(h,0)-f(0,0)}h=\frac{h^5\cdot0}{h(h^4+0^2)}-0=\frac0{h^5}=0\;.$$
As a function of $h$ this is the constant $0$, so its limit as $h\to 0$ is $0$.
By definition: $$\frac{\partial f}{\partial x}(0,0)=\lim_{t \to 0} \frac{f(0+t,0)-f(0,0)}{t} = 0$$ since $f(t,0)=0$ for every $t \in \mathbb{R}$.