0
$\begingroup$

How do I show that the following function is differentiable at $(0,0)$? $$ \begin{cases} \dfrac{\sin(xy)}{y}, & \text{if }y \neq 0 \\ \\ 0, & \text{if }y = 0 \end{cases} $$ I calculated the partial derivatives and

  • $f(x) = \cos(xy)$ exists near $(0,0)$ and is continuous
  • $f(y) = \dfrac{xy \cos(xy) - \sin(xy)}{y^2}$ exists, but how do I show that it is continuous?
$\endgroup$
  • $\begingroup$ If you know power series, $g(u)=\frac{\sin(u)}{u}=\sum_{k\geq 0}(-1)^k\frac{u^{2k}}{(2k+1)!}$ (with $g(0)=1$) is $C^{\infty}$ on $\mathbb{R}$, now your function is $f(x,y)=xg(xy)$, and you are done. $\endgroup$ – Kelenner Jul 2 '17 at 8:05
  • 1
    $\begingroup$ Use the definition of the partial derivative to find $\partial _x f(0,0)$ and $\partial _y f(0,0)$. Then show they are continuous in $(0,0)$. $\endgroup$ – windircurse Jul 2 '17 at 8:10
  • 3
    $\begingroup$ Are you sure that the condition is $(x,y)\neq (0,0)$? Is it maybe $y\neq 0$? $\endgroup$ – Nikolaos Skout Jul 2 '17 at 8:12
2
$\begingroup$

As you defined it your function is not even continuous at $(0,0)$. Note that $$\lim_{y\to0}{\sin(xy)\over y}=x\ .$$ I'm assuming this will be corrected. Then you can argue as follows: When $y\ne0$ you have $${\sin(xy)\over y}=\int_0^x \cos(y t)\>dt\ ,$$ whereby the RHS is obviously $C^1$ in a neighborhood of $(0,0)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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