Partial Differentiation Confusion

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Homework Help Overview

The problem involves finding the partial derivatives of a function \( z \) defined as \( z=\left( [x+y]^3-4y^2 \right)^{\frac{1}{2}} \) with respect to both \( x \) and \( y \). The original poster expresses uncertainty about the form of \( \frac{\partial z}{\partial x} \) compared to \( \frac{\partial z}{\partial y} \).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to differentiate \( z \) with respect to \( y \) and questions whether the same approach applies to \( x \), particularly regarding the treatment of \( y \) as a constant.

Discussion Status

Some participants affirm the original poster's reasoning about keeping \( y \) constant when differentiating with respect to \( x \). There is a shared exploration of the implications of this approach, but no consensus on the final forms of the derivatives has been reached.

Contextual Notes

The discussion includes references to the use of LaTeX for mathematical expressions, indicating a focus on clarity in notation. There is also mention of the educational context in which the original poster is working, specifically related to advanced physics studies.

RazerM
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Homework Statement


Find \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} where z=\left( [x+y]^3-4y^2 \right)^{\frac{1}{2}}

Homework Equations


-

The Attempt at a Solution


I know that \frac{\partial z}{\partial y}=\frac{3(x+y)^2-8y}{2\sqrt{(x+y)^3-4y^2}}
but I am unsure whether \frac{\partial z}{\partial x} is the exact same or does not include the '-8y' in the numerator.

I get the feeling that when finding the derivative inside the square root (in z) that y should still be treated as constant and therefore have no -8y.
 
Last edited:
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Welcome to PF!

Hi RazerM! Welcome to PF! :smile:

(on this forum, you need to type "tex", not "TEX" :wink:)
RazerM said:

Homework Statement


Find \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} where z=\left( [x+y]^3-4y^2 \right)^{\frac{1}{2}}


Homework Equations


-


The Attempt at a Solution


I know that \frac{\partial z}{\partial y}=\frac{3(x+y)^2-8y}{2\sqrt{(x+y)^3-4y^2}}
but I am unsure whether \frac{\partial z}{\partial x} is the exact same or does not include the '-8y' in the numerator.

I get the feeling that when finding the derivative inside the square root (in z) that y should still be treated as constant and therefore have no -8y.

Yes, that's completely correct.

∂z/∂x means "keeping y constant", so that's exactly what you do! :smile:
 


So that means \frac{\partial z}{\partial x}=\frac{3(x+y)^2}{2\sqrt{(x+y)^3-4y^2}}?
 
Yup! :biggrin:

(nice LaTeX, btw :wink:)
 
Thanks :)

I taught myself to use LaTeX to help me with my Physics Investigation as part of Advanced Higher Physics (Highest level of physics taught in school - Scotland), we never got told to use it but no way was I using MS Office or Openoffice's limited equation typesetting, would have been a nightmare :P
 
One of the many benefits of PF membership is that you can now use LaTeX as much as you like! :biggrin:

(in case you haven't found anything similar, a useful bookmark is http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" :wink:)
 
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