Partial differentiation of a function

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SUMMARY

The discussion focuses on the application of partial differentiation to the function z = x^n f(u), where u = y/x. The goal is to demonstrate that x∂z/∂x + y∂z/∂y = nz. Participants emphasize the importance of treating f(u) as a function to be differentiated using the chain rule, specifically noting that f'(u) should be applied while differentiating u with respect to x. This structured approach leads to a clearer understanding of the problem and its solution.

PREREQUISITES
  • Understanding of partial differentiation
  • Familiarity with the chain rule in calculus
  • Knowledge of functions and their derivatives
  • Basic algebraic manipulation skills
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  • Study the chain rule in calculus in more depth
  • Practice problems involving partial differentiation
  • Explore the implications of the product rule in differentiation
  • Learn about implicit differentiation techniques
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Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as educators looking for teaching strategies in partial differentiation.

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Hi, I've got the following problem:

Show that if [tex]z = x^nf(u)[/tex]
and [tex]u = y/x[/tex]
then [tex]x\frac{\partial{z}}{\partial{x}} + y\frac{\partial{z}}{\partial{y}} = nz[/tex]

I know partial differentiation fairly well, but I've never seen one laid out like this before, and am not too sure how to get started (i.e. find [tex]\frac{\partial{z}}{\partial{x}}[/tex]. I don't really know how to treat the [tex]f(u)[/tex]. Any pointers?

Thanks,
Toppers
 
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Treat f(u) like a function to be differentiated using the chain rule. Just write f '(u) and then proceed to differentiating the inside (u) with respect to the x variable (for dz/dx).
 
Thanks snipez, just what I needed. Which was basically to just start writing down the equation and its answers one line after another. Normally with these "show that if x and y then z" questions I can't see the answer coming until the last line, and this was no different.

- Toppers
 

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