- #1
finchie_88
This is the problem word for word out of my textbook:
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.
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:
