Chain Rule and Partial Derivatives for Differentiable Functions

[manenbu]

Homework Statement

Prove that if
z(x,y)=e^y f(ye^{\frac{x^2}{2y^2}})
is differentiable, then
(x^2-y^2) \frac{\partial z}{\partial x} + xy\frac{\partial z}{\partial y} = xyz

Homework Equations

Chain Rule.

The Attempt at a Solution

A similar question is solved like this:
Have this:
z(x,y) = f(x² - y²)

Prove this:
y(∂z/∂x) + x(∂z/∂y) = 0

I define
z = f(u)
then
u = x² - y²
∂z/∂x = ∂z/∂u ∂u/∂x = ∂z/∂u 2x
∂z/∂y = ∂z/∂u ∂u/∂y = ∂z/∂u -2y
adding it up:
y(∂z/∂x) + x(∂z/∂y) = y*∂z/∂u*2x - x*∂z/∂u*2y = 0

When I try the same for the mentioned problem, I run into problems and just can't get it right. I guess it's something with the multipilication in there - but can't figure it out.
What's the correct way?

 

[Office_Shredder]

So for example, to start off
\frac{\partial z}{\partial y} = \frac{\partial (e^y)}{\partial y} f(ye^{\frac{x^2}{2y^2}}) + e^y \frac{\partial f(ye^{\frac{x^2}{2y^2}})}{\partial y}

Why don't you work through the two partial derivatives and let us know what you get?

 

[manenbu]

ok so we got:
\frac{\partial z}{\partial y} = \frac{\partial (e^y)}{\partial y} f(ye^{\frac{x^2}{2y^2}}) + e^y \frac{\partial f(ye^{\frac{x^2}{2y^2}})}{\partial y}
for shortness:
f(ye^{\frac{x^2}{2y^2}}) = f(u) = f
define:
u = ye^{\frac{x^2}{2y^2}}<br />
let's do the derivative:
\frac{\partial z}{\partial y} = e^yf + e^y\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}
noticing that e^yf is the same as z:
\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}
and doing the derivative:
\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}+y\frac{x^2}{2}\frac{-2}{y^3}e^{\frac{x^2}{2y^2}})<br />
let's make it look better:
\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}-\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}) = z + \frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y}(1-\frac{x^2}{y^2})<br />
I'll make this substitution for readbility:
<br /> \frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y} = a<br />
so I end up with:
\frac{\partial z}{\partial y} = z + a(1-\frac{x^2}{y^2})

doing for the other derivative now:
\frac{\partial z}{\partial x} = a\frac{x}{y}
now the proof:
(x^2-y^2)a\frac{x}{y} + xy(z+a-a\frac{x^2}{y^2} = a\frac{x^3}{y}-axy+xya-a\frac{x^3}{y} + xyz = xyz

Problem solved. Thanks a lot!
Was also using this for a bit of tex practice.
Looks so easy now.