Question

What is the derivative of `sin^2 (t/6)`?

Answer 1

`1/3sin(t/6) cos(t/6)`

If sin(t/6) is u then derivative of u^2 w.r.t x would be 2u`(du)/dx`

Now plug in u = `sin(t/6)`, thus required derivative would be `2sin(t/6) d(sin (t/6))/dx`

= `2 sin(t/6) *1/6*cos (t/6)`

=`1/3 sin(t/6) cos(t/6)`