Problem involving partial derivatives.

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SUMMARY

The discussion centers on calculating the differential dz for the function z = (x^3 + y^3) / (x - y) at the points x = 10, y = 8, dx = 2, and dy = -3. The user initially calculated dz as -366 but later confirmed the correct value is -1878 after re-evaluating their partial derivatives. The user derived the partial derivatives correctly but made an error in the final calculation. The correct partial derivatives are ∂z/∂x = 3x^2(x - y)^{-1} - x^3(x - y)^{-2} - y^3(x - y)^{-2} and ∂z/∂y = 3y^2(x - y)^{-1} + y^3(x - y)^{-2} + x^3(x - y)^{-2>.

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finchie_88
This is the problem word for word out of my textbook:

Given that [tex]z = \frac{x^3 + y^3}{x - y}[/tex], x = 10, y = 8, dx = 2, dy = -3, find dz.

Hopefully, someone can tell me where my error(s) are.
This is my method:
[tex]z = \frac{x^3 + y^3}{x - y}[/tex]
[tex]\therefore z = x^3(x-y)^{-1} + y^3(x-y)^{-1}[/tex]
[tex]\frac{\partial{z}}{\partial{x}} = 3x^2(x-y)^{-1} - x^3(x-y)^{-2} - y^3(x - y)^{-2}[/tex] There is a good chance that this is the wrong bit, but I can't see where. I'll show where this came from at the end.

[tex]\frac{\partial{z}}{\partial{y}} = 3y^2(x-y)^{-1} + y^3(x-y)^{-2} + x^3(x - y)^{-2}[/tex]

There is a good chance that this is the wrong bit, but I can't see where. I'll show where this came from at the end.

[tex]\text{Since } dz = \frac{\partial{z}}{\partial{x}}dx + \frac{\partial{z}}{\partial{y}}dy[/tex]

It means that the value of dz given what I have worked out, and given the information given, dz = -366, but the answer is -1878, so where have I gone wrong?

How I got the above pd's:

[tex]\text{Let } z_1 = x^3(x - y)^{-1} => \pd{z_1}{x} = 3x^2(x - y)^{-1} - x^3(x - y)^{-2}[/tex]
[tex]\text{Let } z_2 = y^3(x - y)^{-1}[/tex]
[tex]\frac{\partial{z}}{\partial{x}} = -y^3(x - y)^{-2}[/tex]
[tex]\frac{\partial{z}}{\partial{x}} = \frac{\partial{z_1}}{\partial{x}} + \frac{\partial{z_2}}{\partial{x}}[/tex]
[tex]\frac{\partial{z}}{\partial{x}} = 3x^2(x - y)^{-1} - x^3(x - y)^{-2} - y^3(x - y)^{-2}[/tex]

[tex]\text{Let } z_1 = x^3(x - y)^-1 \text{ then, I got: } \frac{\partial{z_1}}{\partial{x}} = x^3(x - y)^{-2}[/tex]
[tex]\text{Let } z_2 = 3y^2(x - y)^{-1} + y^3(x - y)^{-2} => \frac{\partial{z_2}}{\partial{y}} = -y^3(x - y)^{-2}[/tex]
[tex]\frac{\partial{z}}{\partial{y}} = \pd{z_1}{y} + \pd{z_2}{y}[/tex]
[tex]\frac{\partial{z}}{\partial{y}} = 3y^2(x - y)^{-1} + y^3(x - y)^{-2} + x^3(x - y)^{-2}[/tex]

Thank you, and if you get to this point without getting bored, congratulations! Any kind of help would be kind.:biggrin:

Sorry, it took me a few edits to get the maths readable, even now its not perfect, but oh well.
 
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Frankly, the error must be in your calculation. The derivation is correct. I performed the calculation with the given numbers and got -1878.
 
I can't believe that I made a mistake like that . *sigh*, oh well, thank you for helping me.
 

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