Solving Partial Derivatives: Is This Right?

[Telemachus]

Homework Statement

Hi there. Well, I got the next function, and I'm trying to work with it. I wanted to know if this is right, I think it isn't, so I wanted your opinion on this which is always helpful.

$$f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq-x\\0 & \mbox{si}& y=-x\end{matrix}$$


If $$y\neq-x$$:
$$\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})$$

$$\displaystyle\frac{\partial f}{\partial y}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})$$

Second case:

If $$y=-x$$
$$f(x,y)=0$$

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

Is this right? I would like to know why it isn't, I think its not. Should I think when I consider the second case as I would go in all directions or something like that? cause in other cases where I got defined different functions for a particular point I've used the definition, but in this case I got trajectories. What should I do?

Bye and thanks.

 

[Mark44]

Homework Statement

Hi there. Well, I got the next function, and I'm trying to work with it. I wanted to know if this is right, I think it isn't, so I wanted your opinion on this which is always helpful.

$$f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq-x\\0 & \mbox{si}& y=-x\end{matrix}$$


If $$y\neq-x$$:
$$\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})$$

$$\displaystyle\frac{\partial f}{\partial y}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})$$

Second case:

If $$y=-x$$
$$f(x,y)=0$$

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

Is this right? I would like to know why it isn't, I think its not. Should I think when I consider the second case as I would go in all directions or something like that? cause in other cases where I got defined different functions for a particular point I've used the definition, but in this case I got trajectories. What should I do?

You're not using the product rule correctly in either partial. With ordinary derivative, if h(x) = f(x)*g(x), h'(x) = f'(x)*g(x) + f(x)*g'(x). It's similar for your two partials.

Also, when you differentiate sin(pi/(x + y)), you're not using the chain rule correctly. You will get cos(pi/(x + y)), but now you need the derivative of the argument to the cosine function, so you need the derivative of pi/(x + y).

 

[Telemachus]

Why not?

$$(x+y)^2\sin(\displaystyle\frac{\pi}{x+y})$$

$$\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})$$

The derivative on the left is obvious. For the right part, I've need to find derivative of $$\sin(\displaystyle\frac{\pi}{x+y})$$, which is the derivative of the sine multiplied by the derivative on the inside of the sine (Im working with respect to x):
$$\cos(\displaystyle\frac{\pi}{x+y})(\displaystyle\frac{-\pi}{(x+y)^2})$$
Then, when I make the product with $$(x+y)^2$$ I got $$-\pi\cos(\displaystyle\frac{\pi}{x+y})$$

I don't see what's wrong with it.

 

[Mark44]

You're right. I didn't work the problems through, but it seemed that you were missing a factor on the second terms of your partials. Your work didn't show that intermediate step (before canceling), so I assumed you had not used the chain rule correctly.

 

[Telemachus]

Thanks. Why I have to use the definition of derivative in the second case?

 

[Mark44]

No, I don't see why you would need to use the definition of the derivative. Since f(x, y) = 0 along the line y = -x, both partials are zero as well.