Chain Rule and Partial Derivatives

[theBEAST]

Homework Statement

Here is the problem:
http://dl.dropbox.com/u/64325990/MATH%20253/help.PNG

The Attempt at a Solution

http://dl.dropbox.com/u/64325990/Photobook/Photo%202012-05-24%209%2037%2028%20PM.jpg

This seems to be wrong... Since I have fx and fy which I cannot cancel out. Why is this wrong?

I guess it can only cancel is fx = fy but how do I prove that? We don't even know what f is.

 

[Mark44]

You're more likely to get help if you don't make people open different windows to see the problem and what you did. Speaking for myself only, if you aren't motivated enough to at least try to make it easier for me to help, then I'm likewise not motivated enough to provide any help.

 

[theBEAST]

You're more likely to get help if you don't make people open different windows to see the problem and what you did. Speaking for myself only, if you aren't motivated enough to at least try to make it easier for me to help, then I'm likewise not motivated enough to provide any help.

Oh I'm sorry I just thought it would be easier for the people who are not logged onto their physicsforum account. That way they can see it too. I will upload them to the site as well then...

Edit: I tried to upload the other picture but apparently it was too large. Anyone know how to decrease the image size? I don't have photoshop.

Edit: Ok so turns out you can insert images... But the other image is once again too large. Next time I will edit it on my iphone before uploading it to Dropbox.

 

[Dick]

The partial derivative with respect to x of f(x^2-y^2) is just f'(x^2-y^2)*2x. There's really only one way to take a derivative of f.

 

[theBEAST]

The partial derivative with respect to x of f(x^2-y^2) is just f'(x^2-y^2)*2x. There's really only one way to take a derivative of f.

Thanks but aren't fx and fy different? Since you can take the derivative of f with respect to both x and y. I reuploaded a new attempt I made and it shoes that they are not equal unless fx = fy. But I don't think fx = fy.

 

[Dick]

Thanks but aren't fx and fy different? Since you can take the derivative of f with respect to both x and y. I reuploaded a new attempt I made and it shoes that they are not equal unless fx = fy. But I don't think fx = fy.

f is a function of ONE variable. In your case the value of the variable happens to be x^2-y^2. But there's still only one way to differentiate f. The derivative of f is f'.

 

[theBEAST]

f is a function of ONE variable. In your case the value of the variable happens to be x^2-y^2. But there's still only one way to differentiate f. The derivative of f is f'.

Oh wow that makes so much sense now, thanks!