Finding $\frac{\partial z}{\partial x}$ when sin(5x-4y+z)=0

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Homework Help Overview

The discussion revolves around finding the partial derivative \(\frac{\partial z}{\partial x}\) given the equation \(\sin(5x - 4y + z) = 0\). This involves understanding implicit differentiation and the relationships between the variables involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods for finding \(\frac{\partial z}{\partial x}\), including solving for \(z\) and applying implicit differentiation. There are questions about the implications of treating \(y\) as independent of \(x\) and how to apply the chain rule in this context.

Discussion Status

Some guidance has been offered regarding the use of implicit differentiation and the chain rule. However, there are differing interpretations of the expressions involved, and no consensus has been reached on the correct formulation.

Contextual Notes

Participants are navigating the complexities of implicit differentiation and the assumptions about the independence of variables, which may affect their approaches.

UrbanXrisis
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sin(5x-4y+z)=0

how do I find [tex]\frac{\partial z}{\partial x}[/tex]?

if the problem is sin(5x-4y+z)=f(x,y,z), I can find [tex]\frac{\partial f}{\partial x}[/tex] but I don't know what to do when it is just equal to zero.
 
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Solve for z, take the partial derivative.
 
Simpler: take the partial derivative with respect to x, assuming that y is independent of x, z a function of x, then solve for zx. Use the chain rule. Remember "implicit differentiation" from Calculus I?
(sin(5x- 4y+ z))x= cos(5x- 4y+ z)(5- zx)= 0. Solve for zx.
 
HallsofIvy: I think that should be a (5+zx) in your second expression.
 

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