Question

find the 2nd derivatives of y= 1/x(x^2-3x^2) plz do a solution .. thanks

Answer 1

do you have a CAS?

Answer 2

The answer is -6/(x^5)

Answer 3

what is CAS?

Answer 4

@ziko will u do a solution?

Answer 5

answer is 0
as f'(x) = -2
f''(x) = 0

Answer 6

provide a solution guyz ...

Answer 7

\[y=\frac{ 1 }{ x(x^{2}-3x ^{2} )} \]
or
\[y=\frac{ x ^{2}-3x^{2} }{ x }\]

Answer 8

!!

Answer 9

then what next @ziko1995

Answer 10

Your question is it the derivate of the first function i gaved or the last one !!

Answer 11

really dont understand @ziko1995

Answer 12

@gotyakid which function is it (of the two ziko posted), since y= 1/x(x^2-3x^2) could be either.

Answer 13

i will draw @agent0smith