# Directional derivatives and partial derivatives

• Nov 11th 2012, 09:43 PM
tlawrence
Directional derivatives and partial derivatives
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).

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?
• Nov 12th 2012, 12:09 AM
chiro
Re: Directional derivatives and partial derivatives
Hey tlawrence.

What did you calculate dh/dx and dh/dy in terms of the function f^n(r)? (Note: f^n(r) is the nth derivative of the function f with argument r)?
• Nov 12th 2012, 12:55 AM
tom@ballooncalculus
Re: Directional derivatives and partial derivatives
Also/anyway, f(r) and r(x,y) are differentiable so you can apply the chain rule...

http://www.ballooncalculus.org/draw/diffPartial/six.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x or y), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Similarly with dh/dy. Then substitute into the left-hand side of the show-sentence.

_________________________________________

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Balloon Calculus; standard integrals, derivatives and methods

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