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.