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

##### 1 Answer

#### Explanation:

You can differentiate this function by using implicit differentiation, provided that you don't know what the derivative of

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

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

Focus on finding

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

Use the trigonometric identity

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

to write

#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

#(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))))#