Find all points such that function has all partial derivatives in that point.

Question

Find all points $(x,y) \in \mathbb{R}^2$ such that function has all partial derivatives in that point.$$ f(x,y) = \begin{cases} \frac{\sin(xy^2)}{y} &\mbox{if } y>0 \\ xy^2 & \mbox{if } y \le 0. \end{cases} $$

How to solve this question? I do not know how to even start.

EDIT: Still looking for further explanation

Answer 1

The second sub-function $xy^2$ clearly has all partial derivatives everywhere. The first sub-function $\dfrac{\sin(xy^2)}{y}$ has partial derivatives everywhere it is defined, namely $y\ne 0$: since that sub-function is used only where $y>0$ it clearly has all partial derivatives in that domain.

So the only possible problem is the boundary between those two sub-domains, namely $y=0$. Clearly the partial derivative in $x$ exists everywhere on that line. You need to find where the overall function is continuous there (which it is on the entire line) and where the one-sided derivatives in $y$ are equal.

Any place on the line $y=0$ where the one-sided derivatives in $y$ are not equal is where not all partial derivatives are defined. They are all defined elsewhere.

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Here are more details. We first want to see if $f(x,y)$ is continuous at $y=0$. The value at $y=0$ is $f(x,0)=x\cdot 0^2=0$. To find the left-handed limit we let $x\to x_0$ and $y\to 0^-$ (note that $y<0$ here), and we get

$$\lim f(x,y)=\lim_{x\to x_0,y\to 0-} xy^2=0\cdot x_0\cdot 0^2=0$$

To find the right-handed limit we let $x\to x_0$ and $y\to 0^+$ (note that $y>0$ here), and we get

$$\begin{align} \lim f(x,y) &= \lim_{x\to x_0,y\to 0+} \frac{\sin(xy^2)}{y} \\[2 ex] &=\lim_{x\to x_0}x_0\cdot\lim_{y\to 0^+}y\cdot\lim_{x\to x_0,y\to 0^+}\frac{\sin(xy^2)}{y^2} \\[2 ex] &= x_0\cdot 0\cdot x_0 \quad\text{(since $\lim_{u\to 0}\frac{\sin(au)}{u}=a$)}\\[2 ex] &= 0 \end{align}$$

Since all three values are equal, we see that $f(x,y)$ is indeed continuous on the line $y=0$, hence it is continuous on all of $\Bbb R^2$.

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Now for the partial derivatives. For the left-handed derivative,

$$\begin{align} \frac{\partial f}{\partial y}|_{y=0} &=\frac{d}{dx}(xy^2)|_{y=0} \\ &= 2xy|_{y=0} \\ &= 2x\cdot 0 \\ &= 0 \end{align}$$

For the right-handed derivatives,

$$\begin{align} \frac{\partial f}{\partial y}|_{y=0} &=\frac{d}{dx}\left(\frac{\sin(xy^2)}{y}\right)|_{y=0} \\[2 ex] &=\frac{y\cos(xy^2)\cdot 2xy-\sin(xy^2)\cdot 1}{y^2}|_{y=0} \\[2 ex] &= \left[2x\cos(xy^2)-\frac{\sin(xy^2)}{y^2}\right]|_{y=0} \\[2 ex] &= 2x\cdot 1-x \\[2 ex] &= x \end{align}$$

(Please pardon the slight abuse of notation.) This equals the left-handed partial derivative only when $x=0$.

So your final answer is:

$f(x,y)$ has all partial derivatives at all points except those where ($x\ne 0$ and $y=0$).

Note that this is not two sets of exceptions but one: points where both conditions are true simultaneously. In words, all points have all partial derivatives except for points on the $x$-axis other than the origin.

Answer 2

Hint: $f$ is $C^\infty$ in both $\{y>0\}$ and $\{y<0\},$ so $f$ has partial derivatives of all orders in those domains. So all you have to check is the whether $\partial f \partial x, \partial f \partial y$ exist at points of the $x$-axis.