Partial derivative, I'm not sure if my solution is right

[Telemachus]

Homework Statement

Well, it looks simple, but I'm not sure If the answer I'm giving is right.

The function is:

[tex]f(x,y)=\sin|y|[/tex]
And it asks for the partial derivatives, so:

[tex]\displaystyle\frac{\partial f}{\partial x}=0[/tex]

And
[tex]\displaystyle\frac{\partial f}{\partial y}=\begin{Bmatrix}{ \cos|y|}&\mbox{ if }& y>0\\\not{\exists & \mbox{if}& y=0\\-\cos|y| & \mbox{if}& y<0\end{matrix}[/tex]

Is this right?

 

[rock.freak667]

When you are defining the domains like that, you do not need to rewrite the |y|. But if y=0, then f(x,y)=0 so that ∂f/∂y = 0.

 

[statdad]

"But if y=0, then f(x,y)=0 so that ∂f/∂y = 0."

No. think about [itex] f(x) = |x| [/itex] from ordinary calculus. it is true [itex] f(0) = 0 [/itex] but remember f is not differentiable @ x = 0; similar situation in the case given by the OP

 

[Telemachus]

Its continuous but not differentiable at that point.

 

[statdad]

The absolute value function is continuous at x =0: is your partial derivative continuous in y at y = 0?

 

[Telemachus]

Yes, it is. It can be demonstrated that Lim f(x,y)->(0,0)=f(0,0). So, what you say, the solution I've given is right?

 

[statdad]

yes indeed - i switched my "thinking" to the partial in my previous comment.

 

[Telemachus]

Thanks.