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Math

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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)

  • Math -

    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

  • Math - hmm -

    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))

  • Got me - Math -

    thanks for looking after me,

    rotten copy errors.

  • Math -

    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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