# Chain Rule and Partial Derivatives

1. May 24, 2012

### theBEAST

1. The problem statement, all variables and given/known data
Here is the problem:
http://dl.dropbox.com/u/64325990/MATH%20253/help.PNG [Broken]

3. The attempt at a solution
http://dl.dropbox.com/u/64325990/Photobook/Photo%202012-05-24%209%2037%2028%20PM.jpg [Broken]

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.

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Last edited by a moderator: May 6, 2017
2. May 24, 2012

### Staff: Mentor

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.

3. May 24, 2012

### theBEAST

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.

Last edited: May 24, 2012
4. May 24, 2012

### 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.

5. May 24, 2012

### theBEAST

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.

6. May 24, 2012

### Dick

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'.

7. May 24, 2012

### theBEAST

Oh wow that makes so much sense now, thanks!