How does the chain rule apply to partial derivatives?

[jegues]

Homework Statement

See figure.

Homework Equations

The Attempt at a Solution

Here's what I got,

$$\frac{ \partial z}{\partial x} = \left( \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \right) + \left( \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \right) + \left( \frac{\partial z}{\partial x}\right)$$

Is this correct?

 

[mg0stisha]

I don't know much about partial derivatives, but it seems weird that what you're solving for (dz/dx), is also your last term on the right side of the equivalency.

EDIT: Upon further review, looks like I just don't know enough. Sorry for this not-so-helpful post :(

 

[HallsofIvy]

Yes, that is correct- although since you are given that u= g(x,y) and v= h(x), it would be better to write
$$\frac{\partial z}{\partial x}= \frac{\partial z}{\partial u}\frac{\partial g}{\partial x}+ \frac{\partial z}{\partial v}\frac{dh}{dx}+ \frac{\partial z}{\partial x}$$

 

[Mark44]

HallsofIvy,
That's exactly what I got, but it bothered me that, since $$\frac{\partial z}{\partial x}$$ appears on both sides, there doesn't seem to be any way to get an explicit value for this partial.