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What is the derivative of ##f(theta)=arcsin(sqrt(sin(9theta)))##?
##(df)/(d theta) = 9/2 * cos(9theta)/(sqrt(sin(9theta)) * sqrt(1 - sin(9theta)))##
You can differentiate this function by using , provided that you don't know what the derivative of ##arcsin(x)## is.
You will also need to use the 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 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))))##