Directional derivatives and partial derivatives

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SUMMARY

The discussion focuses on proving the equation x(dh/dx) + y(dh/dy) = rf'(r) for the function h(x,y) = f(√(x² + y²)), where f is differentiable. Participants emphasize the use of the chain rule to derive the partial derivatives ∂f/∂x and ∂f/∂y in terms of r. The key insight is that the relationship involves the limit definition of the derivative and the application of directional derivatives, despite initial confusion about their relevance in this context.

PREREQUISITES
  • Understanding of differentiable functions in calculus
  • Familiarity with the chain rule in multivariable calculus
  • Knowledge of directional derivatives and their definitions
  • Basic proficiency in limits and continuity in calculus
NEXT STEPS
  • Study the application of the chain rule in multivariable calculus
  • Learn about the properties and calculations of directional derivatives
  • Explore the limit definition of derivatives in the context of differentiable functions
  • Investigate the relationship between polar coordinates and Cartesian coordinates in calculus
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus, as well as mathematicians interested in the applications of directional and partial derivatives.

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Homework Statement



Suppose f: R -> R is differentiable and let h(x,y) = f(√(x^2 + y^2)) for x ≠ 0. Letting r = √(x^2 + y^2), show that:

x(dh/dx) + y(dh/dy) = rf'(r)

Homework Equations





The Attempt at a Solution


I have begun by showing that rf'(r) = sqrt(x^2 + y^2) * limt->0 (f(r+t) - f(r))/t

and written out the definition form of the directional derivatives. I can't seem to find a way to equate both sides of the equation. Can anyone help?
 
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I don't see that there is any "directional derivative" involved here. You are given that h(x,y)= f(\sqrt{x^2+ y^2}) . By the chain rule \partial f/\partial y= (df/dr)(\partial r/\partial y) and \partial f/\partial x= (df/dr)(\partial r/\partial x.

With r= \sqrt{x^2+ y^2}, it is easy to find \partial r/\partial x and \partial r/\partial y.
 

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