# Homework Help: Problem involving partial derivatives.

1. Feb 14, 2006

### finchie_88

This is the problem word for word out of my text book:

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

Hopefully, someone can tell me where my error(s) are.
This is my method:
$$z = \frac{x^3 + y^3}{x - y}$$
$$\therefore z = x^3(x-y)^{-1} + y^3(x-y)^{-1}$$
$$\frac{\partial{z}}{\partial{x}} = 3x^2(x-y)^{-1} - x^3(x-y)^{-2} - y^3(x - y)^{-2}$$ 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.

$$\frac{\partial{z}}{\partial{y}} = 3y^2(x-y)^{-1} + y^3(x-y)^{-2} + x^3(x - y)^{-2}$$

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.

$$\text{Since } dz = \frac{\partial{z}}{\partial{x}}dx + \frac{\partial{z}}{\partial{y}}dy$$

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:

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

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

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.

Last edited by a moderator: Feb 14, 2006
2. Feb 14, 2006

### assyrian_77

Frankly, the error must be in your calculation. The derivation is correct. I performed the calculation with the given numbers and got -1878.

3. Feb 15, 2006

### finchie_88

I can't believe that I made a mistake like that:grumpy: . *sigh*, oh well, thank you for helping me.