- #1
issisoccer10
- 35
- 0
[SOLVED] Chain rule problem with partial derivatives
Suppose that z = f(u) and u = g(x,y). Show that..
[tex]\frac{\partial^{2} z}{\partial x^{2}}[/tex] = [tex]\frac{dz}{du}[/tex] [tex]\frac{\partial^{2} u}{\partial x^{2}}[/tex] + [tex]\frac{d^{2} z}{du^{2}}[/tex] [tex]\frac{(\partial u)^{2}}{(\partial x)^{2}}[/tex]
[tex]\frac{\partial z}{\partial x}[/tex] = [tex]\frac{dz}{du}[/tex] [tex]\frac{\partial u}{\partial x}[/tex]
based on the chain rule
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 [tex]\frac{dz}{du}[/tex] and [tex]\frac{\partial u}{\partial x}[/tex] 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 [tex]\frac{\partial u}{\partial x}[/tex] squared, rather than just [tex]\frac{\partial u}{\partial x}[/tex] 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.
Homework Statement
Suppose that z = f(u) and u = g(x,y). Show that..
[tex]\frac{\partial^{2} z}{\partial x^{2}}[/tex] = [tex]\frac{dz}{du}[/tex] [tex]\frac{\partial^{2} u}{\partial x^{2}}[/tex] + [tex]\frac{d^{2} z}{du^{2}}[/tex] [tex]\frac{(\partial u)^{2}}{(\partial x)^{2}}[/tex]
Homework Equations
[tex]\frac{\partial z}{\partial x}[/tex] = [tex]\frac{dz}{du}[/tex] [tex]\frac{\partial u}{\partial x}[/tex]
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 [tex]\frac{dz}{du}[/tex] and [tex]\frac{\partial u}{\partial x}[/tex] 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 [tex]\frac{\partial u}{\partial x}[/tex] squared, rather than just [tex]\frac{\partial u}{\partial x}[/tex] 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.