Derivative help please

1) Calculate the derivative of the function: s(x)= (3x-9/2x+8)^ 3

2) Find the indicated derivative. In this case, the independent variable is a (unspecified) function of t.
s= 3r + r ^ -1 ---> Find ds/dt

3) Compute the indicated derivative using the chain rule: y= 8x+4 ---> Find dx/dy

Please explain. Thank you!

1. Use the chain rule: ds/dx = 3((3x - 9)/(2x + 8)^2 ((2x + 8)3 - (3x - 9)2)/(2x + 8)^2
ds/dx = 3((3x - 9)/(2x + 8)^2)((6x + 24 - 6x + 18)/(2x + 8)^2
ds/dx = 3((3x - 9)/(2x + 8)^2)(42/(2x + 8)^2
ds/dx = 126(3x - 9)^2/(2x + 8)^4

2. ds/dt = 3dr/dt - r^(-2)dr/dt

3. Solve for x: y - 4 = 8x --> y/8 - 1/2 = x --> (1/8) = dx/dy

1. (3x-9/2x+8)^3
This one ain't polite. Chain rule meets quotient
3(3x-9/2x+8)^2 * [(3*2x+8) - (2*(3x-9))]/(2x+8)^2
3(3x-9/2x+8)^2 * [6x+24 - 6x+18]/(2x+8)^2
3(3x-9/2x+8)^2 * 42/(2x+8)^2
126(3x-9)^2 / (2x+8)^4

2. s = 3r +r^-1
This is just applying chain rule. You're using an extra variable that you're not deriving, so you assume it's an entire function all it's own.
s' = 3r' -r^-2 r'
s' = 3r' - r'/r^2

3. y = 8x + 4
y = 8*(x)^1 + 4
y' = 8* (1*x^0 * 1)
y' = 8* (1)
y' = 8

1) this will use the chain rule, along with the power rule and quotient rule
s (x) = [3(x - 3) / [2(x + 4)]]^3 = (3/2)^3 [(x - 3) / (x + 4)]^3 = 27/8 [(x - 3) / (x + 4)]^3

s ' (x) = 27/8 * 3[(x - 3) / (x + 4)]^2 [(x + 4) - (x - 3)] / (x + 4)^2
s ' (x) = 81 / 8 [(x - 3)(7)] / (x + 4)^4
s ' (x) = 567 / 8 (x - 3) / (x + 4)^4

*****
2) s = 3r + r^(-1)
ds / dt = 3 dr/dt - r^(-2) dr/dt = (3 - r^(-2)) dr/dt

*****
y = 8x + 4
dy/dx = 8
therefore, dx/dy = 1/8

or, dy = 8 dx ==> dxy/dy = 1/8