Directional derivatives and partial derivatives

  • Thread starter The1TL
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  • #1
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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 cant seem to find a way to equate both sides of the equation. Can anyone help?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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I don't see that there is any "directional derivative" involved here. You are given that [itex]h(x,y)= f(\sqrt{x^2+ y^2})[/itex] . By the chain rule [itex]\partial f/\partial y= (df/dr)(\partial r/\partial y)[/itex] and [itex]\partial f/\partial x= (df/dr)(\partial r/\partial x[/itex].

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

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