Show that $f(x, y) := e^{xy} \sin(x + y)$ is totally differentiable

Asked Jun 30 '16 at 10:31

Active May 14 '18 at 21:48

Viewed 142 times

Given,

$$f(x, y) := e^{xy} \sin(x + y),$$

I want to show that $f(x, y)$ is totally differentiable.

Approach

Since we were never given any example for solving a problem like this, I feel pretty much lost. I'd guess that I have to evaluate the gradient of $f$ at some point, so that's what I already did:

$$\nabla f = {ye^{xy} \sin(x + y) + e^{xy}\cos(x + y) \choose xe^{xy} \sin(x + y) + e^{xy} \cos(x + y)}$$

Unfortunately, I don't see where to go from here.

Edit after discussion in the comments

I have to show that the partial derivatives are continiuous - which they are since they are compositions of continiuous functions. Therefore, $f$ is totally differentiable.

edited Jun 12 '20 at 10:38

Community

asked Jun 30 '16 at 10:31

Julian

1,401

You've basically done it... do those partial derivatives exist everywhere? – Merkh Jun 30 '16 at 10:46
2

Just check that the partial derivatives are continuous; this is a well-known sufficient condition for differentiability. – Hans Lundmark Jun 30 '16 at 10:50
Is this definitely the same procedure for totally differentiability? – Julian Jun 30 '16 at 10:53
Ah, I found the necessary theorem. I have to show that the partial derivatives are continiuous - as you said. :-) I'll try to figure that out. – Julian Jun 30 '16 at 10:58
Okay, well, I guess it's kind of obvious that the partial derivatives are continiuous since they are a composition of continiuous functions, aren't they? – Julian Jun 30 '16 at 11:04
@Noobian: yes indeed – Juan Fran Jun 30 '16 at 12:20
Forget about the "totally" here. The question is whether this function is "differentiable". – Christian Blatter Jun 30 '16 at 14:53
Why? It was mentioned in the excercise. Or are "totally differentiable" and "differentiable" the same properties? – Julian Jun 30 '16 at 15:05

Show that $f(x, y) := e^{xy} \sin(x + y)$ is totally differentiable

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