Math
posted by Ifi .
How to d/dx
y= (9+((x9)/6)^1/5)^1/2 ?
Hope u can give me the way how to calculate it. ^^
I got 1/60 ((x9)/6) ^ (9/10), i think my answer is wrong because i check at wolfram website the ans is differrent from mine (sorry i didn't write the ans bcus the ans is quite long and difficult to type)

not quite ....
let u = ( (x9)/6 )^(1/5)
then y = (9 + u)^1/2
dy/du = (1/2)(9+u)^(1/2)
du/dx = (1/5)( (x9)/6 )^(4/5) (1/6)
= (1/30) ( (x9)/6 )^(4/5)
so dy/dx = dy/du * du/dx
= (1/2)(9 + u)^1/2) (1/30) ( (x9)/6 )^(4/5)
= (1/60) (9 + ( (x9)/6 )^(1/5)^(1/2) ( (x9)/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 + ( (x9)/6 )^1/2 , then by implicit differentiation,
2y dy/dx = (1/2)( (x9)/6) )^(1/2)
dy/dx = (1/4)(√(6/(x9) ) / y
or √6/(4y√(x9) ) , that's not bad 
not bad, but
y^2 = 9 + ((x9)/6)^(1/5)
2y dy/dx = 1/5 ((x9)/6)^(4/5)
dy/dx = 1/(10y ((x9)/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 + ((x9)/6)^(1/5)
2y dy/dx = 1/5 ((x9)/6)^(4/5)) (1/6)
dy/dx = 1/60y ((x9)/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)?