What is the derivative of #f(theta)=arcsin(sqrt(sin(9theta)))#?

You can differentiate this function by using implicit differentiation, provided that you don't know what the derivative of #arcsin(x)# is.

You will also need to use the chain rule and the fact that

#d/dx(sinx) = cosx#

Your starting function

#f(theta) = arcsin(sqrt(sin(9theta)))#

is equivalent to

#sin(f) = sqrt(sin(9theta)) " "color(orange)((1))#

Differentiate both sides with respect to #theta#

#d/(d theta)(sin(f)) = d/(d theta)sqrt(sin(9theta))#

Focus on finding #d/(d theta)sqrt(sin(9theta))# first. Use the chain rule twice to get

#d/(d theta)sqrt(sin(9theta)) = 1/2sin(9theta)^(-1/2) * cos(9theta) * 9#

#d/(d theta)sqrt(sin(9theta)) = 9/2 sin(9theta)^(-1/2) * cos(9theta)#

Take this back to your target derivative to get

#cos(f) * (df)/(d theta) = 9/2sin(9theta)^(-1/2) * cos(9theta)#

Isolate #(df)/(d theta)# on one side

#(df)/(d theta) = 9/2 * (sin(9theta)^(-1/2) * cos(9theta))/cos(y)#

Use the trigonometric identity

#color(blue)(sin^2x + cos^2x = 1)#

to write #cosx# as a function of #sinx#

#cos^2x = 1 - sin^2x implies cosx = sqrt(1- sin^2x)#

This will get you

#(df)/(d theta) = 9/2 * (sin(9theta)^(-1/2) * cos(9theta))/sqrt(1-sin^2(y))#

Finally, use equation #color(orange)((1))# to get

#(df)/(d theta) = 9/2 * (sin(9theta)^(-1/2) * cos(9theta))/sqrt(1- (sqrt(sin(9theta)))^2)#

This is equivalent to

#(df)/(d theta) = color(green)(9/2 * cos(9theta)/(sqrt(sin(9theta)) * sqrt(1 - sin(9theta))))#