Question

How to d/dx
y= (9+((x-9)/6)^1/5)^1/2 ?
Hope u can give me the way how to calculate it. ^^

I got 1/60 ((x-9)/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)

Answer 1

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

Answer 2

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))

Answer 3

thanks for looking after me,

rotten copy errors.

Answer 4

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)?