# Taking Partial Derivatives

1. Sep 2, 2008

### Jacobpm64

1. The problem statement, all variables and given/known data
If $$z = ax^2 + bxy + cy^2$$ and $$u = xy$$, find $$\left(\frac{\partial z}{\partial x}\right)_{y}$$ and $$\left(\frac{\partial z}{\partial x}\right)_{u}$$ .

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

3. 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.

2. Sep 2, 2008

### HallsofIvy

Staff Emeritus
Replace y in the function by u/x.

3. Sep 2, 2008

### 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?