# Implicit Differentiation (for a normal derivative) with Multivariable Functions

1. Nov 11, 2011

### paul2211

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

z$^{3}$x+z-2y-1=0
xz+y-x$^{2}$+5=0

Define z as a function of x, find z'.

2. Relevant equations

I guess the two equations above...

3. The attempt at a solution

Well, I just differentiated the first one with respect to x and got:

3z$^{2}$xz'+z$^{3}$3+1=0
z' = $\frac{-1-z^{3}{3z$^{2}$x}$

If I differentiate the second equation, I get something completely different...

z+xz'-2x=0
z' = $\frac{2x-z}{x}$

I'm sorry if I typed some of my work above wrong, but the main reason I'm stuck on this question is: I don't even see why I need the two equations to solve this problem

Thanks in advance to you guys.

Last edited: Nov 11, 2011
2. Nov 11, 2011

### I like Serena

Welcome to PF, paul2211!

Your variable y is a function of x and z.
As such you need to consider it as well when differentiating.

Note that in this case you can use your second equation to eliminate y altogether.

Btw, you appear to have made at least one type when differentiating the first equation.

3. Nov 11, 2011

### paul2211

So does this mean y = y(x, z(x))?

If so, and I somehow couldn't cancel the y between the two equations, wouldn't I have $\frac{\partial y}{\partial x}$, $\frac{\partial y}{\partial z}$ and $\frac{dz}{dx}$ to solve for?

Since I would only have two equations, does this mean I can't solve for $\frac{dz}{dx}$ if I didn't have the insight to cancel out the other variable? (y in this case).

4. Nov 11, 2011

### I like Serena

Yes and yes.

If you didn't cancel y beforehand, y will probably cancel itself as long as you apply the proper derivative rules.
It's just less work to cancel y beforehand.