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 September 1st, 2016, 05:05 AM #1 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 Chain rule and second order derivatives I'm trying to work out the following question:- Use the chain rule to show that (d2y/dx2).((dx/dy)^3) + d2x/dy2 = 0 I picked y=x^3 to convince myself that it is correct (for that equation) but don't know how to go about using the chain rule to prove it There is a hint saying: (dy/dx).(dx/dy) = 1 differentiate this Thanks
 September 1st, 2016, 05:14 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra What happens if you write $v(y)={\mathrm d x \mathrm d y}$, giving $y'v=1$? Differentiate implicitly. Thanks from topsquark and wirewolf
 September 1st, 2016, 04:15 PM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 dy/dx=y' dx/dy=1/y' d$\displaystyle ^{2}$x/dy$\displaystyle ^{2}$=-(1/y'$\displaystyle ^{2}$)y''(dx/dy)=-(1/y'$\displaystyle ^{3}$)y'' Substituting into OP: y''(1/y'$\displaystyle ^{3}$)-(1/y'$\displaystyle ^{3}$)y''=0 Thanks from topsquark
September 2nd, 2016, 05:45 AM   #4
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Quote:
 Originally Posted by v8archie What happens if you write $v(y)={\mathrm d x \mathrm d y}$, giving $y'v=1$? Differentiate implicitly.
Nothing happens. A function can't equal a product of differentials.

 September 10th, 2016, 03:13 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Yes. There was quite obviously a typo in that post. Luckily the OP was able to see that, even if you were not.
 September 10th, 2016, 06:50 PM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 There was no indication anyone understood what you meant until after I responded.
 September 10th, 2016, 07:47 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra You mean apart from the people who thanked me.
September 10th, 2016, 08:09 PM   #8
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Quote:
 Originally Posted by wirewolf There is a hint saying: (dy/dx).(dx/dy) = 1 differentiate this
OP thanked you after my post. If they understood what you meant, why would they thank you for repeating the hint in OP?

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