# Partial Differentiation Question

1. Nov 16, 2014

1. The problem statement, all variables and given/known data
if $z=\frac{1}{x^2+y^2-1}$ . Show that $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = -2z(1+z)$

2. Relevant equations
n/a

3. The attempt at a solution
I am extremely new to partial differentiation, I can get my head around questions where they just give you the function and ask for the partial wrt to any of the variables but have not come accross a problem like this and am a little unsure of the notation, does $x \frac{\partial z}{\partial x}$ litterally mean x multipled with the partial wrt x ?

First what I did was to find the partial derivatives the question asks for so done them below.
$$\frac{\partial z}{\partial x} = \frac{-2}{x^3+y^2-1} \\ \frac{\partial z}{\partial y} = \frac{-2}{x^2+y^3-1} \\$$

So if it just means x multiplied by.... then wouldn't it just become...
$$x \frac{\partial z}{\partial x} = \frac{-2x}{x^3+y^2-1} \\ y \frac{\partial z}{\partial y} = \frac{-2y}{x^2+y^3-1} \\$$

Or do I need to solve for x and y and substitute that into the above for x and y?

Last edited: Nov 16, 2014
2. Nov 16, 2014

### CAF123

Yes, the notation $x \frac{\partial z}{\partial x}$ means multiply the partial derivative by $x$. So you have the right idea but you haven't computed $\partial z/\partial x$ or $\partial z/\partial y$ correctly.

3. Nov 16, 2014

### Staff: Mentor

Yes, that's all it means.
Both of the above are wrong. It's easier to use the chain rule with z written as (x2 + y2 - 1)-1.

4. Nov 16, 2014

### Zondrina

Your derivatives look wrong though. Lets find $\frac{\partial z}{\partial x} = z_x$ first.

I suggest you re-write $z$ like so:

$$z = (x^2 + y^2 - 1)^{-1}$$

Now what is $z_x$? Don't forget the chain rule!

5. Nov 16, 2014

Ah right ok thanks, will try again and post in a minute or so

6. Nov 16, 2014

Ok so using the chain rule for $\frac{\partial z}{\partial x}$

if $h=x^2+y^2-1$ and $m=h^{-1}$

Then
$$h' \cdot m' = 2x \cdot -h^{-2} = -2x(x^2+y^2-1)^{-2}$$

And for $\frac{\partial z}{\partial y}$
$$-2y(x^2+y^2-1)^{-2}$$

EDIT: Oh and is $z_x$ a short hand notation for $\frac{\partial z}{\partial x}$ then? As it does get a bit annoying writing them out all the time.

7. Nov 16, 2014

### Zondrina

Yes. Your derivatives look okay now too.