Math Help - Partial derivative using chain rule. Can't find my mistake.

1. Partial derivative using chain rule. Can't find my mistake.

Evaluate $\frac{\partial w}{\partial u}$ at (u,v)=(1,3) for the function

$w(x,y,z)=xz+yz-z^{2};x=uv,y=uv,z=u$

Ok, so I'm going to to do:

$(\frac{\partial w}{\partial x})(\frac{dx}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial y})(\frac{dy}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})$

Right?

Then I got:

$(z)(v)+(x)(1)+(z)(v)+(y)(1)-(2z)(2u)$

After substituting:

$=>zv+x+zv+y-4uz$

$=>uv+uv+uv+uv-4u^{2}|_{(1,3)}$

$=>(1)(3)+(1)(3)+(1)(3)+(1)(3)-4(1^{2})=8$

But this isn't one of the answer choices on my exam review. What did I do wrong?

2. Originally Posted by downthesun01
Evaluate $\frac{\partial w}{\partial u}$ at (u,v)=(1,3) for the function

$w(x,y,z)=xz+yz-z^{2};x=uv,y=uv,z=u$

Ok, so I'm going to to do:

$(\frac{\partial w}{\partial x})(\frac{dx}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial y})(\frac{dy}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})$ Mr F says: What are the last two terms doing there?

Right?

Then I got:

$(z)(v)+(x)(1)+(z)(v)+(y)(1)-(2z)(2u)$ Mr F says: Please explain clearly where each of these terms have come from. Some of them make no sense at all to me.

[snip]
..