Partial Differentiation Confusion

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.
 
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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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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