Please check my 2d derivative

[hank]

For f(x) = (x-2)^3 / x^2
find f'(x) and f"(x).

Here's what I get:
f'(x) = 3x^2(x-2)^2 - 2x(x-2)^3
---------------------------------------
x^4

= x(x-2)^2(3x-2x+4)
-----------------------
x^4

f'(x) = (x-2)^2 * (x + 4)
---------------------
x^3

= (x^3) - 12x + 16
------------------
x^3

f"(x) = (3x-12)x^3 - (x^3 - 12x+16)3x^2
----------------------------------------
x^6

= x(3x-12) - 3(x^3 - 12x + 16)
-----------------------------------
x^4

= 3x^2 - 12x - 3x^3 + 36x -48
-----------------------------------
x^4

Does this look right?
I feel pretty good about my first derivative, but I'm not liking my 2d derivative at all.

 

[arthur ohlsten]

y=[x-2]^3 / x^2 an improper fraction. I find it easiest to change to non fractional form. Expand numerator:
y= [x-2][x^2-4x+4] / x^2 expand by clearing paranthesis
y=[x^3-6x^2+12x-8]/x^2 rewrite
y=x-6+12x^-1-8x^-2 take derivative

y'=1-12x^-2 +16x^-3 --- or--- [x^3-12x+16]/x^3 take derivative

y''= 24x^-3-48x^-4 ---- or ---- 24[x-2]/x^4

Arthur

 

[hank]

Ok, I see where I made my mistake.

Thank you!