Posted by Ifi on Saturday, October 27, 2012 at 10:25am.
not quite ....
let u = ( (x-9)/6 )^(1/5)
then y = (9 + u)^1/2
dy/du = (1/2)(9+u)^(-1/2)
du/dx = (1/5)( (x-9)/6 )^(-4/5) (1/6)
= (1/30) ( (x-9)/6 )^(-4/5)
so dy/dx = dy/du * du/dx
= (1/2)(9 + u)^-1/2) (1/30) ( (x-9)/6 )^(-4/5)
= (1/60) (9 + ( (x-9)/6 )^(1/5)^(-1/2) ( (x-9)/6 )^(-4/5)
I suggest you check my steps for any errors, I should have written it out on paper first.
Wolfram seems to have performed some kind of complicated simplification.
How about this way:
square both sides:
y^2 = 9 + ( (x-9)/6 )^1/2 , then by implicit differentiation,
2y dy/dx = (1/2)( (x-9)/6) )^(-1/2)
dy/dx = (1/4)(√(6/(x-9) ) / y
or √6/(4y√(x-9) ) , that's not bad
not bad, but
y^2 = 9 + ((x-9)/6)^(1/5)
2y dy/dx = 1/5 ((x-9)/6)^(-4/5)
dy/dx = 1/(10y ((x-9)/6)^(4/5))
thanks for looking after me,
rotten copy errors.
thanks for the calculation but i think there is a mistake at implicit diff.
y^2= 9 + ((x-9)/6)^(1/5)
2y dy/dx = 1/5 ((x-9)/6)^(-4/5)) (1/6)
dy/dx = 1/60y ((x-9)/6)^(-4/5) ..... Sorryif im wrong
Btw why at wolfram the answer is
1/( 60y(-9+y^2)^4). Why is the x become y^2 and why there are no 6 and ^(4/5)?
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