This is the problem word for word out of my text book:(adsbygoogle = window.adsbygoogle || []).push({});

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.

Sorry, it took me a few edits to get the maths readable, even now its not perfect, but oh well.

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# Homework Help: Problem involving partial derivatives.

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