College Calculus 1

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We have to find the first and second derivative of f(x)=x^(2/3)(6-x)^(1/3)
I have the first derivative as being
(4-x)/(x^(1/3)(6-x)^(2/3))
And that I know is right. The second derivative is -8/(x^(4/3)(6-x)^(5/3))
I did all the work but I am not getting the right answer... I looked it over but I don't know where I am going wrong... can someone do like the fisrt several lines? Or the whole thing... Thanks!

  • College Calculus 1 -

    Is there anyone that can help me with this!!!!!!!!!!!!!! PLEASE!!!!!!

  • College Calculus 1 -

    d/dx((6-x)^(1/3) x^(2/3))
    | Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = (6-x)^(1/3) and v = x^(2/3):
    = | x^(2/3) (d/dx((6-x)^(1/3)))+(6-x)^(1/3) (d/dx(x^(2/3)))
    | Use the chain rule, d/dx((6-x)^(1/3)) = ( du^(1/3))/( du) ( du)/( dx), where u = 6-x and ( du^(1/3))/( du) = 1/(3 u^(2/3)):
    = | (x^(2/3) (d/dx(6-x)))/(3 (6-x)^(2/3))+(6-x)^(1/3) (d/dx(x^(2/3)))
    | The derivative of x^(2/3) is 2/(3 x^(1/3)):
    = | (x^(2/3) (d/dx(6-x)))/(3 (6-x)^(2/3))+(2 (6-x)^(1/3))/(3 x^(1/3))
    | Differentiate the sum term by term and factor out constants:
    = | (x^(2/3) (d/dx(6)-d/dx(x)))/(3 (6-x)^(2/3))+(2 (6-x)^(1/3))/(3 x^(1/3))
    | The derivative of 6 is zero:
    = | (2 (6-x)^(1/3))/(3 x^(1/3))-(x^(2/3) (d/dx(x)))/(3 (6-x)^(2/3))
    | The derivative of x is 1:
    = | (2 (6-x)^(1/3))/(3 x^(1/3))-x^(2/3)/(3 (6-x)^(2/3))

    d^2/dx^2(x^(2/3) (6-x)^(1/3)) = -(2 (6-x)^(1/3))/(9 x^(4/3))-(2 x^(2/3))/(9 (6-x)^(5/3))-4/(9 (6-x)^(2/3) x^(1/3))

  • College Calculus 1 -

    The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is the initial height in feet (as if you were on top of a tower or building).

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