Calculus

posted by Mike

Differentiate and simplify as much as possible.

Cube root(5z+6/-9z+3).

The answer should be y'=
(23(-9z+23)^2/3)/9(3z-1)^2(5z+6)^2/3)

So far, I'm stuck at y'=[23/3(5z+6)(3z+1)][((5z+6)^1/3)/((-9z+3)^1/3)]

  1. Steve

    if you let w be the fraction, then you have

    w^(1/3)

    which has derivative

    1/3 w^(-2/3) dw/dz

    Now just use the quotient rule on dw/dz, and you get

    69/(-9z+3)^2

    That gives you

    1/3 * ((5z+6)/(-9z+3))^(-2/3) * 69/(-9z+3)^2

    1 * (-9z+3)^(2/3) * 69
    ---------------------------------------
    3 * (5z+6)^(2/3) * (-9z+3)^2

    which you can see is the desired answer. I see a (3z+1) in your answer, which is already a typo. Try breaking it up into factors as I did, and you should see things cancel.

  2. Reiny

    using the quotient rule, by first line derivative is

    y' = (1/3)[(5z+6)/(3-9z)]^(-2/3) [ (5(3-9z) - (-9)(5z+6) ]/(3-9z)^2
    = (1/3)[(5z+6)/(3-9z)]^(-2/3) [ 69/(3-9z)^2 ]
    = 23 (3-9z)^(2/3) / (5z+6)^2/3 [ 1/(3-9z)^2 ]
    = not getting your answer

    let's try the product rule
    y = (5z+6)^(1/3) ( 3-9z)^(-1/3)

    y' = (5z+6)^(1/3) (-1/3)(3-9z)^(-4/3) (-9) + ( 3-9z)^(-1/3) (1/3)(5z+6)^(-2/3) (5)
    = (9/3)(5z+6)^(1/3) (3-9z)^(-4/3) + (5/3)(3-9z)^(-1/3) (5z+6)^(-2/3)
    = (1/3)(5z+6)^(-2/3) (3-9z)^(-4/3) [ 9(5z+6) + 5(3-9z)]
    = 23(5z+6)^(-2/3) (3-9z)^(-4/3)

    check through my steps

  3. Reiny

    looks like Steve and I have the same answer

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