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))))##