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if y'=(1/x') therefore x'=(1/y')   so d(5)y/dx(5) =y''''' and x'''''=1/[d(5)y/dx(5)]=dx(5)/d(5)y

Posted on 2016-09-06
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Last Modified: 2016-10-10
I had this question after viewing Is (dx²)/(d²y) is acceptable as second derivative?.
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Question by:John Sebastian
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by:aburr
ID: 41786375
Your equation has difficulties with the meaning that you attach to it. Consider your premise

if y'=(1/x')

It is wrong. Try it on a simple function like y = x^3. It will not work.     (If you have trouble, ask)
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by:John Sebastian
ID: 41789244
y=x^3

y'=3x²

1=3x²x'   →x'=1/3x²

y'=1/x' is correct
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by:aburr
ID: 41789959
1=3x²x'  
I have a hard time figuring out how you got this relationship.
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by:John Sebastian
ID: 41790326
Dx/dy
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by:aburr
ID: 41790525
but if y = x^3      x = y^(1/3)
dx/dy is then =
                                (1/3)y^(-2/3)      not 3x²x'
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Author Comment

by:John Sebastian
ID: 41790782
1=3x²x'→x'=1/3x²→x'=1/3(y⅓
→x'=1/3(y⅔)
Note that x=y^(1/3)
Therefore they have the same answer... Am i wrong?
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by:aburr
ID: 41791935
"Am i wrong?"   no    I am
I put one too many inverses in my calculations
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BigRat earned 500 total points (awarded by participants)
ID: 41808671
>>y'=1/x' is correct

That is ALWAYS correct simply from the definition of y' (as you have it) since y' = dy/dx and x'=dx/dy and dx/dy*dy/dx=1.

Similarly if you want to write y'' for d²y/dx² then x'' would be d²x/dy² and so y''=x''. And so on and so forth PROVIDED that such differentials exist!
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by:aburr
ID: 41836561
The only correct answer, (and a good one)
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