Partial Derivatives: Show bz(x)=az(y)

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SUMMARY

The discussion centers on the problem of demonstrating that bz(x) = az(y) given the function z = f(ax + by), where a and b are constants. The initial attempt incorrectly assumes z = ax + by, leading to the conclusion that bz(x) = az(y) through direct differentiation. The correct approach involves applying the chain rule, recognizing that u = ax + by is a more general representation of z, which allows for proper differentiation with respect to x and y.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with the chain rule in calculus
  • Knowledge of functions and their compositions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the application of the chain rule in multivariable calculus
  • Explore the properties of partial derivatives in functions of multiple variables
  • Investigate examples of differentiating composite functions
  • Learn about the implications of constant coefficients in partial differentiation
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable functions and partial derivatives, as well as educators looking for examples to illustrate the chain rule in action.

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Homework Statement



Suppose that z=f(ax+by), where a and b are constants. Show that bz(x) = az(y).

z(x) means partial derivative of z with respect to x, as for z(y).

Homework Equations





The Attempt at a Solution



Say z=ax+by

z(x) = a

z(y) = b

So bz(x) = ba = ab = az(y)

I'm not sure that this is a correct analysis - because I know f(ax+by) doesn't necessarily mean z = ax+by... How should I interpret this problem?
 
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Keep f general and use the chain rule. It might help to let u = ax + by.
 

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