# Math Help - relative extrema

1. ## relative extrema

use both the first and second derivative test to show that f(x)=sin^2 x has a relative minimum of x=0

i keep getting that x=0 has no min or max because it equals zero in the second derivative test this is what im doing see if you can see what im doing wrong

f '(x)=2cosx and i set that equal to zero and i get x=0 then
f ''(x)=-sinx which is equal to zero when i plug x in and that gives me no min or max

2. Originally Posted by vinson24
use both the first and second derivative test to show that f(x)=sin^2 x has a relative minimum of x=0

i keep getting that x=0 has no min or max because it equals zero in the second derivative test this is what im doing see if you can see what im doing wrong

f '(x)=2cosx and i set that equal to zero and i get x=0 then
f ''(x)=-sinx which is equal to zero when i plug x in and that gives me no min or max
You have to use the chain rule here:

$f(x)=\sin^2(x)~\implies~f'(x)=2\sin(x)\cdot \cos(x)$