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$f(x)=\frac{(y')^2}{x^3}$

Find $\frac{d}{dx} \frac{\partial f}{\partial y'}$

I don't understand how to take this derivative properly. Can someone describe step by step?

kiwifruit
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  • You need to clarify. What is $y'$? – Christopher K Mar 25 '14 at 03:27
  • y' comes from the functional: $J(x, y')=\int_0^2 \frac{(y')^2}{x^3}dx$ – kiwifruit Mar 25 '14 at 03:32
  • If you're using the Euler-Lagrange equation, you should be working with $\partial / \partial y'$ not $\partial / \partial y$. – user137500 Mar 25 '14 at 03:34
  • So you should probably disregard my 'answer', which I don't think makes any sense anyway. – user137500 Mar 25 '14 at 03:35
  • Oh, sorry, yes, I corrected the typo in the question. Still having trouble with this derivative. – kiwifruit Mar 25 '14 at 03:35
  • Incidentally, do you need any help solving the differential equation?! :)) – user137500 Mar 25 '14 at 03:46
  • That would be useful, too - got the derivative and I know it must be equal to 0, but no idea how to solve it. Also, I have another calculus of variation question with a derivative/solution confusion and an open bounty if you are interested - http://math.stackexchange.com/questions/722753/function-extremal-calculus-of-variations – kiwifruit Mar 25 '14 at 03:49

1 Answers1

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damn it now that's all wrong... Try 2:

$$\frac{\partial}{\partial y'}\frac{(y')^2}{x^3}=\frac{2y'}{x^3}$$ and $$\frac{d}{dx}\left(\frac{2y'}{x^3}\right)=\frac{2y''x^3-3x^2(2y')}{x^6}$$ I'll make you simplify since it got all messed up the first time :)

user137500
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