College Calculus 1
posted by Please help .
We have to find the first and second derivative of f(x)=x^(2/3)(6x)^(1/3)
I have the first derivative as being
(4x)/(x^(1/3)(6x)^(2/3))
And that I know is right. The second derivative is 8/(x^(4/3)(6x)^(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!

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

d/dx((6x)^(1/3) x^(2/3))
 Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = (6x)^(1/3) and v = x^(2/3):
=  x^(2/3) (d/dx((6x)^(1/3)))+(6x)^(1/3) (d/dx(x^(2/3)))
 Use the chain rule, d/dx((6x)^(1/3)) = ( du^(1/3))/( du) ( du)/( dx), where u = 6x and ( du^(1/3))/( du) = 1/(3 u^(2/3)):
=  (x^(2/3) (d/dx(6x)))/(3 (6x)^(2/3))+(6x)^(1/3) (d/dx(x^(2/3)))
 The derivative of x^(2/3) is 2/(3 x^(1/3)):
=  (x^(2/3) (d/dx(6x)))/(3 (6x)^(2/3))+(2 (6x)^(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 (6x)^(2/3))+(2 (6x)^(1/3))/(3 x^(1/3))
 The derivative of 6 is zero:
=  (2 (6x)^(1/3))/(3 x^(1/3))(x^(2/3) (d/dx(x)))/(3 (6x)^(2/3))
 The derivative of x is 1:
=  (2 (6x)^(1/3))/(3 x^(1/3))x^(2/3)/(3 (6x)^(2/3))
d^2/dx^2(x^(2/3) (6x)^(1/3)) = (2 (6x)^(1/3))/(9 x^(4/3))(2 x^(2/3))/(9 (6x)^(5/3))4/(9 (6x)^(2/3) x^(1/3)) 
The equation S= 16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where âvâ is the initial velocity (speed) in ft/sec and âkâ is the initial height in feet (as if you were on top of a tower or building).
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