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September 1st, 2016, 05:05 AM   #1
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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

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September 1st, 2016, 05:14 AM   #2
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What happens if you write $v(y)={\mathrm d x \mathrm d y}$, giving $y'v=1$? Differentiate implicitly.
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September 1st, 2016, 04:15 PM   #3
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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
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September 2nd, 2016, 05:45 AM   #4
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Quote:
Originally Posted by v8archie View Post
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.
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September 10th, 2016, 03:13 PM   #5
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Yes. There was quite obviously a typo in that post. Luckily the OP was able to see that, even if you were not.
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September 10th, 2016, 06:50 PM   #6
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There was no indication anyone understood what you meant until after I responded.
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September 10th, 2016, 07:47 PM   #7
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You mean apart from the people who thanked me.
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September 10th, 2016, 08:09 PM   #8
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Quote:
Originally Posted by wirewolf View Post
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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