Chain rule problem with partial derivatives

[issisoccer10]

[SOLVED] Chain rule problem with partial derivatives

Homework Statement

Suppose that z = f(u) and u = g(x,y). Show that..

$$\frac{\partial^{2} z}{\partial x^{2}}$$ = $$\frac{dz}{du}$$ $$\frac{\partial^{2} u}{\partial x^{2}}$$ + $$\frac{d^{2} z}{du^{2}}$$ $$\frac{(\partial u)^{2}}{(\partial x)^{2}}$$

Homework Equations

$$\frac{\partial z}{\partial x}$$ = $$\frac{dz}{du}$$ $$\frac{\partial u}{\partial x}$$

based on the chain rule

The Attempt at a Solution

Based on the first order partial derivative above, I would think that using the product rule we can find the second order partial dervative of z w.r.t. x

Using my intuition, I consider $$\frac{dz}{du}$$ and $$\frac{\partial u}{\partial x}$$ like different terms and then apply the product rule.

However, I know this isn't correct because I am supposed to show that in the last term of the question equation we have $$\frac{\partial u}{\partial x}$$ squared, rather than just $$\frac{\partial u}{\partial x}$$ as I would conclude.

If my attempted solution doesn't make any sense, I'll try to clarify. But it is wrong either way and any help in finding the correct way to get the desired equation would be greatly appreciated.

 

[Dick]

It might be clearer if you write the whole thing in terms of f and g. But you are on the right track, it's product rule and chain rule put together. When you take another d/dx derivative of dz/du you will get the second derivative of z together with yet another du/dx in addition to the one you've already got, giving you a (du/dx)^2. Do you see that, or do I need to tex it?

 

[issisoccer10]

oh ok...so

the derivative of $$\frac{dz}{du}$$ = $$\frac{d^{2} z}{du^{2}}$$ $$\frac{\partial u}{\partial x}$$

because we really are taking the derivative with respect to x.. so basically there was another chain that I didn't see..

is that right?

 

[Dick]

oh ok...so

the derivative of $$\frac{dz}{du}$$ = $$\frac{d^{2} z}{du^{2}}$$ $$\frac{\partial u}{\partial x}$$

because we really are taking the derivative with respect to x.. so basically there was another chain that I didn't see..

is that right?

I think so. There's a chain you didn't see all right. If you get what you were supposed to then you are doing it right.

 

[issisoccer10]

thanks a lot..

 

[H.Geologist]

Would anyone tex the solution in a step by step form please?
Thank you in advance