Calculus help!!please...answer correctly get 10 points?

Question

Evaluate dy/dx at the specified value of x:
1. y=6(cuberoot of x + 2)^2 ,x=8
2. y=x^3+4x^-1 ,x=1

what is the correct answers?help

Answer 1

Answer-1.
y=6(cuberoot of x + 2)^2

dy/dx = 6*2(cuberoot of x + 2)*(1/3)*(x^((1*3) -1)
dy/dx = 4(x^((1*3) +2)*(x^(-2/3))
dy/dx(x=8) = 4(8^((1*3) +2)*(8^(-2/3))
dy/dx(x=8) = 4(2 +2)*(8^(-2/3))
dy/dx(x=8) = 4(4)*(1/4))
dy/dx(x=8) = 4

Answer-2
y=x^3+4x^-1 ,x=1

dy/dx = 3*x^2 + 4*(-1)x^(-2) -0
dy/dx(x=1) = 3*(1^2) +4* (-1)1^(-2)
dy/dx(x=1) = 3*(1) - 4
dy/dx(x=1) = -1

I hope thats ok

Answer 2

y=6((x+2)^2)^(1/3) =6(x+2)^(2/3)
dy/dx = (2/3)(6)(x+2)^(-1/3)
dy/dx= 4(8+2)^(-1/3) = 4/)cuberoot (10) at x = 8

y=x^3 + 4x^-1
dy/dx = 3x^2 -4x^-2 3x^2-4/x^2
= 3(1^2) -4/(1^2)= 3-4 =-1 at x =1

Answer 3

1)

y'=12((x^(1/3))+2)*(1/3)x^(-2/3)
at x=8; y'=4

2)

y'=3(x^2) - 4(x^(-2))
at x=1; y'=-1