# Homework Help: Partial derivatives

1. Mar 4, 2012

### bobsmith76

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

In the steps below, the ∂z/∂x does not seem to be obeying normal algebraic rules. I'm confused. This is not really a problem, I'm just trying to understand the steps.

3. The attempt at a solution

1. 3z2∂z/∂x - y + y∂z/∂x = 0
2. ∂z/∂x = y/(y + 3z2)

if ∂z/∂x were a normal algebraic variable, say p, it would be

2p = y/(y + 3z2)

so why not

2∂z/∂x = y/(y + 3z2)

2. Mar 4, 2012

### Fredrik

Staff Emeritus
I don't see how you got that extra factor of 2.
\begin{align}&3z^2 p - y + yp = 0\\
&(3z^2+y)p = y\\
&p =\frac{y}{3z^2+y}
\end{align}

3. Mar 4, 2012

### Staff: Mentor

bobsmith,
What Fredrik did in his 2nd step was to factor out p (AKA ∂z/∂x), essentially doing the opposite of the distributive property. This is yet another example of where weakness in your algebraic knowledge is preventing you from understanding very simple operations. Until and unless you rectify this problem, you will continue to be unable to make sense of what you're reading.