Solving for ∂z/∂x: Partial Derivatives Confusion

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SUMMARY

The discussion centers on the confusion surrounding the calculation of the partial derivative ∂z/∂x in the equation 3z²∂z/∂x - y + y∂z/∂x = 0. The correct solution is derived as ∂z/∂x = y/(y + 3z²) after factoring out ∂z/∂x. A critical point highlighted is the misunderstanding of algebraic manipulation, specifically the application of the distributive property in this context. The discussion emphasizes the importance of solid algebraic foundations to grasp such mathematical operations effectively.

PREREQUISITES
  • Understanding of partial derivatives and their notation
  • Familiarity with algebraic manipulation techniques
  • Knowledge of the distributive property in algebra
  • Basic calculus concepts, particularly in multivariable functions
NEXT STEPS
  • Study the principles of partial differentiation in multivariable calculus
  • Review algebraic manipulation techniques, focusing on factoring and distribution
  • Practice solving partial derivative problems using different functions
  • Explore resources on common pitfalls in calculus to strengthen foundational knowledge
USEFUL FOR

Students studying calculus, particularly those struggling with partial derivatives, educators teaching multivariable calculus, and anyone looking to reinforce their algebraic manipulation skills.

bobsmith76
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Homework Statement



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.

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)
 
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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}
 
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.
 

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