Derivative of a quotient of functions

[Kamhogo]

Hello. I'm stuck with a differential problem. I checked my work over and over but I don't get the same answer as the textbook.
My answer: -141; the textbook's: -5.64 .Can someone please tell me what I'm doing wrong?

S = (2n^2 - 3n + 8)/(2 - 3n^4)

dS/dn = [(2n - 3n^4)((6n^2) -3) - (2n^3 - 3n + 8)(2 -12n^3)]/(2n- 3n^4)^2

= [ 12n^3 - 6n - 18n^6 + 9n^4 -( 4n^3 - 24n^6 - 6n + 36n^4 + 16 - 96n^3)]/ (2n - 3n^4)^2

= ( 12n^4 - 6n - 18n^6 + 9n^4 - 4n^3 + 24n^6 + 6n - 36n^4 - 16 + 96n^3)/ (2n - 3n^4)^2

= (6n^6 - 27n^4 + 104n^3 - 16)/(2n - 3n^4)^2

If x = -1, then dS/dn = (6 - 27 - 104 - 16)/ ( -2 + 3)^2

= -141 / 1 = -141

 

[stapel]

S = (2n^2 - 3n + 8)/(2 - 3n^4)

dS/dn = [(2n - 3n^4)((6n^2) -3) - (2n^3 - 3n + 8)(2 -12n^3)]/(2n- 3n^4)^2

Is there maybe a typo or two in the original exercise statement? Please confirm or correct. Thank you!

 

[Kamhogo]

No, there's no typo. I think I found my mistake. At the end, I made a mistake, where I subsituted n with -1:

(6n^6 - 27n^4 + 104n^3 -16) / ( 2n - 3n^4)^2 =( 6-27-104-16) /(-2 + 3 ) = -141/1= - 141

Correct answer: (6-27-104-16)/(-2-3)^2 = -141/(-5)^2= -141/25 = -5.64

Am I correct?

 

[Kamhogo]

Oh yes, there's a typo. Correction:
S= ( 2n ^3 - 3n + 8 ) / ( 2n - 3n^4)
dS/dn = [(2n - 3n^4)(6n^2 -3) - (2n^3 - 3n + 8)(2 - 12n^3)]/ (2n - 3n^4)^2

= (6n^6 - 27n^4 + 104n^3 -16)/(2n-3n^4)^2

Substituting n with -1:

= (6-27-104-16)/(-2-3)^2 = -141 / ( - 5 )^2 = -141/25 = -5.64

Correct?

And thanks a lot !