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Implicit Differentiation (for a normal derivative) with Multivariable Functions

  • Thread starter paul2211
  • Start date
  • #1
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Homework Statement



z[itex]^{3}[/itex]x+z-2y-1=0
xz+y-x[itex]^{2}[/itex]+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[itex]^{2}[/itex]xz'+z[itex]^{3}[/itex]3+1=0
z' = [itex]\frac{-1-z^{3}{3z[itex]^{2}[/itex]x}[/itex]

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

z+xz'-2x=0
z' = [itex]\frac{2x-z}{x}[/itex]

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:

Answers and Replies

  • #2
I like Serena
Homework Helper
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Welcome to PF, paul2211! :smile:

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
36
1
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 [itex]\frac{\partial y}{\partial x}[/itex], [itex]\frac{\partial y}{\partial z}[/itex] and [itex]\frac{dz}{dx}[/itex] to solve for?

Since I would only have two equations, does this mean I can't solve for [itex]\frac{dz}{dx}[/itex] if I didn't have the insight to cancel out the other variable? (y in this case).
 
  • #4
I like Serena
Homework Helper
6,577
176
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.
 

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