# Thread: I cannot solve this...More partial derivatives...

1. ## I cannot solve this...More partial derivatives...

I have read the book, tried looking at similar examples, but I simply cannot solve the following problem:

The function f(u,v) is differentiable in the entire R^2. Write:

h(x,y,z)= f(x/y, y/z)

y>0, z> 0
(do not understand why this is of importance)

Calculate:

x*(dh/dx)+y*(dh/dy)+z*(dh/dz) expressed in u, v and partial derivates of f.

The dh/dx, dh/dy and dh/dz all have those crooked d's which I cannot type here.

I really need help with this and would be so thankful for any guidance!

2. $\frac {\partial h}{\partial x} \: = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial x} =\frac {\partial f}{\partial u}\frac {1}{y}$

$where$

$\frac {\partial u}{\partial x} =\frac {\partial (x/y)}{\partial x} =1/y$

$\frac {\partial v}{\partial x} =\frac {\partial (y/z)}{\partial x} =0$

As

$u=x/y \: it \: follows \: that \: y \ne 0$

3. Thanks, but should I do the calculation for dh/dy and dh/dz as well...?

I get:

dh/dy= (df/du)*(-x/y^2)+(df/dv)*(1/z)
dh/dz= (df/dv)*(-y/2)

Shall I mulitply this with x, y and z respectively? The key answer says:
x*(dh/dx)+y*(dh/dy)+z*(dh/dz)= 0
But I dont understand how...

Help would be gold...

4. $
x\frac {\partial h}{\partial x}=\frac {x}{y}\frac {\partial f}{\partial u}
$

$y\frac {\partial h}{\partial y}=...$

$z\frac {\partial h}{\partial z}=...$

and make sum of three terms.