Question 1156572
{{{F(theta) = arcsin(sqrt(sin(3theta)))}}}

apply the chain rule : 

{{{df(u )/dx=(df/du)*(du/dx)}}} where

{{{f= arcsin(u)}}}
{{{u=sqrt(sin(3theta))}}}


=>{{{((d/du)(arcsin(u)))*((du/dx)(sqrt(sin(3theta))))}}}

then

{{{(d/du) (arcsin(u))=1/(sqrt(1-u^2))}}}

and

{{{(du/dx)(sqrt(sin(3theta)))= 3cos(3theta)/(2sqrt(sin(3theta)))}}}


so,

=>{{{((d/du)(arcsin(u)))*((du/dx)(sqrt(sin(3theta))))=(1/(sqrt(1-u^2)))(3cos(3 theta)/(2sqrt(sin(3theta))))}}}


substitute back{{{ u=sqrt(sin(3theta))}}}

{{{F}}}'{{{(theta) =(1/(sqrt(1-(sqrt(sin(3theta)))^2)))(3cos(3 theta)/(2sqrt(sin(3theta))))}}}

{{{F}}}'{{{(theta) =(1/(sqrt(1-sin(3theta))))(3cos(3theta)/(2sqrt(sin(3theta))))}}}......simplfy

{{{F}}}'{{{(theta) = (3 cos(3theta))/(2sqrt(-(sin(3theta) - 1) sin(3theta)))}}}