I have a problem with this function.
I have to study where the function is continuous and where is differentiable.
The function is $f : \Bbb R^2 → \Bbb R$ :
$$
f(x, y) = \begin{cases}
x + \sin(y)& x\le y \\
y + \sin(x) & x>y \\
\end{cases}
$$
**Continuity**
(i think that is wrong):
I know that
$$ sin(y) + x $$
is continuos in $\mathbb{R}^2$ because $sin(y)$ and $x$ are continuos.
(for $sin(x)+ y$ is the same)
My problem is about the point $(x = y)$.
What i have to do?
I tried with:
$$
\lim_{(x,y)\to (x,x)}f(x,y)
$$
**Differentiability**
I compute the Partial derivative:
$$
\frac{\partial f}{\partial x} = \begin{cases}
1& x\le y \\
cos(x) & x>y \\
\end{cases}
$$
and
$$
\frac{\partial f}{\partial y} = \begin{cases}
cos(y)& x\le y \\
1 & x>y \\
\end{cases}
$$
So i think that is differentiable in $\mathbb{R}^2$\\{(x,y): x=y}.
What i have to do now?
i tried to use the definition of partial derivative without results...
I have to study where the function is continuous and where is differentiable.
The function is $f : \Bbb R^2 → \Bbb R$ :
$$
f(x, y) = \begin{cases}
x + \sin(y)& x\le y \\
y + \sin(x) & x>y \\
\end{cases}
$$
**Continuity**
(i think that is wrong):
I know that
$$ sin(y) + x $$
is continuos in $\mathbb{R}^2$ because $sin(y)$ and $x$ are continuos.
(for $sin(x)+ y$ is the same)
My problem is about the point $(x = y)$.
What i have to do?
I tried with:
$$
\lim_{(x,y)\to (x,x)}f(x,y)
$$
**Differentiability**
I compute the Partial derivative:
$$
\frac{\partial f}{\partial x} = \begin{cases}
1& x\le y \\
cos(x) & x>y \\
\end{cases}
$$
and
$$
\frac{\partial f}{\partial y} = \begin{cases}
cos(y)& x\le y \\
1 & x>y \\
\end{cases}
$$
So i think that is differentiable in $\mathbb{R}^2$\\{(x,y): x=y}.
What i have to do now?
i tried to use the definition of partial derivative without results...