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.

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?

  • 1
    $\begingroup$ Ok. Any attempts or thoughts? $\endgroup$ – Rebellos May 3 '20 at 8:21
  • $\begingroup$ I have to remove the absolute value in order to differentiate. I approach the problem when x+y < 0 and when x + y >= 0 and it turns out that the function is the same in both cases because sin(-a) = -sin(a). So the derivatives can be easily found and they exist everywhere. It seems to be quite the obvious approach but I have no idea how to describe in the language of mathematics. $\endgroup$ – Sinestro White May 3 '20 at 8:33
  • $\begingroup$ When the question wants you to show the existence of derivatives, you have to work by the definition. Also, since you had a thought process, why not include all that in your original post? $\endgroup$ – Rebellos May 3 '20 at 8:35
  • $\begingroup$ I will, sure, give me a few minutes. $\endgroup$ – Sinestro White May 3 '20 at 8:38
  • $\begingroup$ The method of "removing the absolute value in order to differentiate" only works for points away from the line $y=-x$. You must handle the case $y=-x$ separately. $\endgroup$ – Leander Tilsted Kristensen May 3 '20 at 8:42

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.

  • $\begingroup$ What do you find wrong about this $ \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$ when $x+y=0$ $\endgroup$ – Sinestro White May 3 '20 at 14:42
  • $\begingroup$ That it is not true that $f(x + h, y) - f(x,y) = 0 + h - h$ for every $h$ near $0$. In fact for $h \ne 0$, it isn't true for any $|h| < \pi/2$ (or most $h$ greater than $\pi/2$, but there are exceptions on that side). $\endgroup$ – Paul Sinclair May 3 '20 at 15:04

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 $x=-y.$

  • $\begingroup$ How can you differentiate when you have absolute value? @Allawonder $\endgroup$ – Sinestro White May 3 '20 at 14:49
  • $\begingroup$ @SinestroWhite Yes, except from the points where the argument of the absolute value function vanishes; hence the last clause in my answer above. $\endgroup$ – Allawonder May 3 '20 at 15:06

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