Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $$f(x,y) = \begin{cases} \frac{x^2 y^2}{x^4 + y^4}, & \text{ }\text{(x,y)} \neq (0,0) \\ 0, & \text{ }\text{(x,y)} = (0,0) \end{cases} .$$

Show that $\frac{df}{dx} (0,0)$ exists, and $f$ is not continuous at $(0,0)$.

share|improve this question
1  
Write down the limit defining the partial derivative. What do you get? Compute the limit of $f$ along the path $x=y$ and compare with that value of $f(0,0)$. What do you get? –  David Mitra Apr 2 '12 at 23:27
add comment

1 Answer 1

up vote 2 down vote accepted

For existence of the partial derivative:

By definition, $f_x(0,0)$ is the limit $$ \lim_{h\rightarrow 0}{f(h,0)-f(0,0)\over h }. $$ Show that this limit exists for your particular function.

For continuity:

Note that,
$$ \lim_{x\rightarrow 0} f(x,x) =\lim_{x\rightarrow 0}{x^4\over2x^4}={1\over2}. $$ Keeping this in mind and the fact that $f(0,0)=0$, what can you say about the continuity of $f$ at $(0,0)$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.