Show that partial derivatives exist at every point and that they are continuous

Question

I have a problem with the following question:

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ and $f(x,y) = |x+y|\sin|x+y|$

a) Show that the function $f(x,y)$ has derivatives at every point $(x,y) \in \mathbb{R}^2$.
b) Find the points where the partial derivatives are continuous.
c) Find the points where the function $f(x,y)$ is differentiable.

EDIT:
What I have managed to do so far:

When $x + y > 0$

$f(x,y) = (x + y)sin(x+y)$

$ \displaystyle{\frac{\partial f}{\partial x}} =\displaystyle{\lim_{h \to 0}} \frac{f(x+h,y)-f(x,y)}{h} =$

$ \displaystyle{\lim_{h \to 0}} \frac{(x+y+h)sin(x+y+h)-(x+y)sin(x+y)}{h} =$



$ \displaystyle{\lim_{h \to 0}} \frac{(x+y+h)sin(x+y+h) - (x+y+h)sin(x+y) }{h}+ \displaystyle{\lim_{h \to 0}} \frac{ (x+y+h)sin(x+y) -(x+y)sin(x+y)}{h}=$



$ \displaystyle{\lim_{h \to 0}} \frac{(x+y+h)(sin(x+y+h) - sin(x+y)) }{h}$
$+ \displaystyle{\lim_{h \to 0}} \frac{sin(x+y)((x+y+h) - (x+y))}{h}=$



$ \displaystyle{\lim_{h \to 0}}(x+y+h).\displaystyle{\lim_{h \to 0}} \frac{(sin(x+y+h) - sin(x+y)) }{h}$
$+ sin(x+y).\displaystyle{\lim_{h \to 0}} \frac{((x+y+h) - (x+y))}{h}=$



$ (x+y).\displaystyle{\lim_{h \to 0}} \frac{(sin(x+y+h) - sin(x+y)) }{h}$
$+ sin(x+y).\displaystyle{\lim_{h \to 0}} \frac{((x+y+h) - (x+y))}{h}=$



$ (x+y).\displaystyle{cos(x+y)}$
$+ sin(x+y).\displaystyle{\lim_{h \to 0}} \frac{h}{h}=$



Finally, we get:

$\displaystyle{\frac{\partial f}{\partial x}} = (x+y).\displaystyle{cos(x+y)}+ sin(x+y)$



Let's find the other derivative:

$ \displaystyle{\frac{\partial f}{\partial y}} =\displaystyle{\lim_{h \to 0}} \frac{f(x,y+h)-f(x,y)}{h} = \displaystyle{\lim_{h \to 0}} \frac{(x+y+h)sin(x+y+h)-(x+y)sin(x+y)}{h}$

So we get the same result as for $\frac{\partial f}{\partial x}$
$\displaystyle{\frac{\partial f}{\partial y}} = (x+y).\displaystyle{cos(x+y)}+ sin(x+y)$



When $x + y < 0$

$f(x,y) = -(x + y)sin(-(x+y)) \Rightarrow f(x,y) = -(x + y)(-sin(x+y)) \Rightarrow $

$ f(x,y) = (x + y)sin(x+y)$

Consequently the partial derivatives are going to be the same in this case.



When $x + y = 0$

$f(x,y) = 0.sin(0) \Rightarrow f(x,y) = 0$

$ \displaystyle{\frac{\partial f}{\partial x}} =\displaystyle{\lim_{h \to 0}} \frac{f(x+h,y)-f(x,y)}{h} =\displaystyle{\lim_{h \to 0}} \frac{(0+h)-h}{h} =0$

$\displaystyle{\frac{\partial f}{\partial y}} = 0$

Is this enough to show that there are partial derivatives at every point $(x,y) \in \mathbb{R}^2$? The partial derivatives seem to be continues everywhere but how do I put this in terms of mathematics?

Answer 1

Note that $f(x,y) = h(g(x,y))$ where $h(t) = |t|\sin |t|$ and $g(x,y) = x + y$. When taking the partial derivatives, use the chain rule on this composition.

Answer 2

You have that $$\mathrm df=|x+y|\mathrm d\sin|x+y|+\sin|x+y|\mathrm d|x+y|=|x+y|\cos|x+y|\mathrm d|x+y|+\sin|x+y|\mathrm d|x+y|=(|x+y|\cos|x+y|+\sin|x+y|)\mathrm d|x+y|=(|x+y|\cos|x+y|+\sin|x+y|)\frac{x+y}{|x+y|}\mathrm d(x+y)=\left((x+y)\cos|x+y|+\frac{x+y}{|x+y|}\sin|x+y|\right)(\mathrm dx+\mathrm dy),$$ for every point not on the line $x=-y.$