0
$\begingroup$

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?

$\endgroup$
14
  • 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
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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 $x=-y.$

$\endgroup$
2
  • $\begingroup$ 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

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.