What are the first and second derivatives of the function f(x)=3x/2x^2-18

I assume this is

3x /   [ 2x^2 - 18 ]        if so, we can wrtite this as

(3x) ( 2x^2 - 18)^(-1)

First derivative

3 ( 2x^2 - 18)^(-1)  - 3x (2x^2 - 18)^(-2) (4x) =

3( 2x^2 - 18 )^(-1) - 12x^2 (2x^2 - 18)^(-2)   =

3 ( 2x^2 - 18)^(-2)    [ (2x^2 - 18 - 4x^2 ] =

3 (2x^2 - 18)^(-2) [ -2x^2 - 18 ] =

3 ( 2x^2 - 18)^(-2) [ -2 (x^2 + 9 ] =

-6 (x^2 + 9) (2x^2 - 18)^(-2)

Second derivative

- 6  [   2x (2x^2 - 18)^(-2)  - 2 (x^2 + 9) (2x^2 - 18)^(-3) (4x) ]   =

-6 [  2x (2x^2 - 18)^(-2)   - 8x (x^2 + 9) (2x^2 - 18)^(-3) ]  =

-6 [ 2x (2x^2 - 18)^(-3)  ] [   (2x^2 - 18) - 4( x^2 + 9) ]   =

-6 [ 2x (2x^2 - 18)^(-3) ]   [ 2x^2 - 18 - 4x^2 - 36 ]   =

-6 [ 2x (2x^2 - 18)^(-3) [ -2x^2 - 54 ]   =

6 [ 2x (2x^2 - 18)^(-3) (2)(x^2 + 27) ]  =

12 [ 2x (2x^2 - 18 )^(-3) (x^2 + 27)   =

24x (x^2 + 27) / [ (2^3 (x^2 - 9)^(-3) ] =

24x  (x^2 + 27) / [ 8 ( x^2 - 9)^(-3) ] \

3x (x^2 + 27) (x^2 - 9 )^(-3)