# Partial Differentiation

1. Oct 8, 2005

### elle

Hi, can anyone help me with the following differentiation question?

Find first partial derivatives w.r.t to x and y for:

tan (x/y)

Can anyone offer any help on how to approach this question? I know when you differentiate (x/y) using the product rule, that you have to differentiate y implicitly...but how does that work when using partial derivatives?

Thanks for taking time to read

2. Oct 8, 2005

### mezarashi

In partial derivatives, you treat all variables as constants (you know, the a, b, c)... except for the one you are differentiating with respect to.

For example:

f(x,y,z) = 2xyz + 3x + 4y + 5z
df/dz = 2xy + 5
df/dx = 2yz + 3
df/dy = 2xz + 4

3. Oct 9, 2005

### elle

Ok, I've given it a go :uhh: but I'm not sure if its right...

f (x,y) = tan (x/y)

Let f(x,y) = tan (u) where u = (x/y) = xy^(-1) *** Not sure if I've used the right method for this part

f1 (w.r.t x) = (1/y) sec^2 (x/y)

f2 (w.r.t y) = -(x/y^2) sec^2 (x/y)

Hmm can someone let me know if its right? :uhh:

4. Oct 9, 2005

### arildno

Yes, you have gotten the correct answers.

5. Oct 9, 2005

### elle

Thank you! Glad to know Im on the right track

If I had to use partial differentiation for (1/2) ln (x^2 + y^2), can I just use the same technique (chain rule) and:

Let (1/2) ln (x^2 + y^2) = 1/2 ln (u) where u = x^2 + y^2

Then:

f1 (w.r.t x) = x ln (x^2 + y^2)

f2 (w.r.t y) = y ln (x^2 + y^2) ?

6. Oct 9, 2005

### arildno

Certainly you can do that.

7. Oct 9, 2005

### elle

ok thanks! There is just one last part of da question regarding the logs which I don't understand.

Show that x f(1) + y (f2) = 1?

I've started with:

x^2 ln (x^2 + y^2) + y^2 ln (x^2 + y^2)

so how do I show that it is equal to 1?

8. Oct 9, 2005

### HallsofIvy

Staff Emeritus
I don't understand this. Is "x f(1)+ y(f2)= 1" supposed to be
xfx+ yfy= 1?

Are you saying that f(x,y)= x2ln(x2+y2)+ y2ln(x2+y2)?

If those are what you mean, then you can't show that
xfx+ yfy= 1, it's not true.

9. Oct 9, 2005

### arildno

LEARN PROPER NOTATION!!!!!
:grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy:
Now, f(1) as you call it should be $$\frac{\partial{f}}{\partial{x}}=\frac{x}{x^{2}+y^{2}}$$ and f(2) is really $$\frac{\partial{f}}{\partial{y}}=\frac{y}{x^{2}+y^{2}}$$
when $$f(x,y)=\frac{1}{2}ln(x^{2}+y^{2})$$

10. Oct 9, 2005

### elle

Yes it is, I dunno how to do the superscript x you see

Erm I've attached the question here, though its suppose to be 'u' not 'f'

http://tinypic.com/eg5mhi.jpg

11. Oct 9, 2005

### arildno

Just click on the Latex images to see how the code is for the various symbols.