Implicit Differentiation (for a normal derivative) with Multivariable Functions

[paul2211]

Homework Statement

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

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

Homework Equations

I guess the two equations above...

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.

 

[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.

 

[paul2211]

Thank you for your quick reply Serena.

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).

 

[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.