What is the Method for Finding Partial Derivatives with Extra Functions?

[Jacobpm64]

Homework Statement

If z = ax^2 + bxy + cy^2 and u = xy, find \left(\frac{\partial z}{\partial x}\right)_{y} and \left(\frac{\partial z}{\partial x}\right)_{u} .

Homework Equations

I have Euler's chain rule, the "splitter" and the "inverter" for dealing with partial derivatives.

The Attempt at a Solution

I think finding \left(\frac{\partial z}{\partial x}\right)_{y} is easy.
\left(\frac{\partial z}{\partial x}\right)_{y} = 2ax + by

However, I do not know how to begin to find \left(\frac{\partial z}{\partial x}\right)_{u} because of the extra function u. One thought is substituting u for xy in the second term on the right side of the original equation ( i wouldn't know how to differentiate it though).

Any kind of direction would be helpful.

Thanks in advance.

 

[HallsofIvy]

Homework Statement

If z = ax^2 + bxy + cy^2 and u = xy, find \left(\frac{\partial z}{\partial x}\right)_{y} and \left(\frac{\partial z}{\partial x}\right)_{u} .

Homework Equations

I have Euler's chain rule, the "splitter" and the "inverter" for dealing with partial derivatives.

The Attempt at a Solution

I think finding \left(\frac{\partial z}{\partial x}\right)_{y} is easy.
\left(\frac{\partial z}{\partial x}\right)_{y} = 2ax + by

However, I do not know how to begin to find \left(\frac{\partial z}{\partial x}\right)_{u} because of the extra function u. One thought is substituting u for xy in the second term on the right side of the original equation ( i wouldn't know how to differentiate it though).

Any kind of direction would be helpful.

Thanks in advance.

Replace y in the function by u/x.

 

[Jacobpm64]

All right, let me try your suggestion:

z = ax^2 + bx\left(\frac{u}{x}\right) + c\left(\frac{u}{x}\right)^2
z = ax^2 + bu + \frac{cu^2}{x^2}

\left(\frac{\partial z}{\partial x}\right)_{u} = 2ax - \frac{2cu^2}{x^3}

I guess I can resubstitute to get:
\left(\frac{\partial z}{\partial x}\right)_{u} = 2ax - \frac{2cy^2}{x}

Is this correct?

Thanks in advance.