# Chain Rule and Partials

1. Sep 4, 2009

### manenbu

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Chain Rule.

3. 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
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?

Last edited: Sep 4, 2009
2. Sep 4, 2009

### Office_Shredder

Staff Emeritus
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?

3. Sep 4, 2009

### 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:
$$u = ye^{\frac{x^2}{2y^2}}$$
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 eyf 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:
$$\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}+y\frac{x^2}{2}\frac{-2}{y^3}e^{\frac{x^2}{2y^2}})$$
let's make it look better:
$$\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}-\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}) = z + \frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y}(1-\frac{x^2}{y^2})$$
I'll make this substitution for readbility:
$$\frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y} = a$$
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.