Question

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

Answer 1

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

Explanation:

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))))`