Partial Derivatives: Find Pdz/Pdu & Pdz/Pdv

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Homework Help Overview

The discussion revolves around finding partial derivatives of a function Z defined in terms of variables x and y, which are themselves expressed in terms of u and v. The specific derivatives being explored are ∂z/∂u and ∂z/∂v, with participants working through the relationships between these variables.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to apply the chain rule for partial derivatives, calculating ∂z/∂u and ∂z/∂v based on the relationships provided. Questions arise regarding the correctness of their calculations, particularly in the signs of the terms in their final expressions.

Discussion Status

Some participants confirm the calculations presented, while others point out potential errors in the signs of the terms. The discussion appears to be productive, with participants actively engaging in verifying each other's work and clarifying their understanding of the derivative calculations.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance they can provide to one another. There is also a focus on ensuring the correct application of mathematical principles without providing complete solutions.

brendan
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Homework Statement



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

Homework Equations



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

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
 
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Hi Brendan! :smile:

(have a curly d: ∂ and try using the X2 tag just above the Reply box :wink:)
brendan said:
Let Z = 3x-2y x = u+v ln(u) and y = u^2-v ln(v)

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?

Yes that's fine! :smile:

(except for the minuses in the last line :wink:)
 
Thanks a lot for your help.
regards
Brendan
P.S what does the X^2 tag do?
 
brendan said:
P.S what does the X^2 tag do?

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

To find ∂z/∂v

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

Yes that's fine! :smile:

(except for the minuses in the last line … again :rolleyes:)
 
Thanks once again!
 

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