# Partial derivatives

1. Mar 29, 2009

### brendan

1. The problem statement, all variables and given/known data

Let Z = 3x-2y x = u+v ln(u) and y = u^2-v ln(v)

2. Relevant equations

find Pdz/Pdu and Pdz/Pdv Pd (partial Derivative)

3. The attempt at a solution

Pdz/Pdx = 3 Pdz/Pdy = -2

Pdx/Pdu = v/u +1 Pdx/Pdv = ln(u)

Pdy/Pdu = 2u Pdy/Pdv = -ln(v)-1

Pdz/Pdu = Pdz/Pdx * Pdx/Pdu + Pdz/Pdy * Pdy/Pdu

= 3 * (v/u +1) + -2 * 2u

=3(u+v)/u + (-4u)

= (-4u^2 - 3u - 3v)/u

Could someone please let me know if this is correct?
regards
Brendan

2. Mar 30, 2009

### tiny-tim

Hi Brendan!

(have a curly d: ∂ and try using the X2 tag just above the Reply box )
Yes that's fine!

(except for the minuses in the last line )

3. Mar 30, 2009

### brendan

regards
Brendan
P.S what does the X^2 tag do?

4. Mar 30, 2009

### tiny-tim

If you click the tag, it types [noparse], and anything in between goes above the line and gets smaller …

so you see yx2n in the Reply box, but in the post you see[/noparse] yx2n

5. Mar 30, 2009

### brendan

For the same equation,

Let Z = 3x-2y x = u+v ln(u) and y = u^2-v ln(v)

To find ∂z/∂v

∂z/∂v = ∂z/∂x * ∂x/∂v + ∂z/∂y * ∂y/∂v

We have ∂z/∂x = 3, ∂x/∂v = ln(u), ∂z/∂y= -2, and ∂y/∂v= -ln(v)-1

Is it:

3*ln(u)-2*(ln(v)+1) ?

regards
Brendan

6. Mar 30, 2009

### tiny-tim

Yes that's fine!

(except for the minuses in the last line … again )

7. Mar 30, 2009

### brendan

Thanks once again!