What is the derivative of y = ln(sin^2(theta))

I didn't get it quite right on my test (got half credit), but I was wondering if someone could walk me through it. Thanks!

I'll use t instead of theta.

y = ln(sin^2(t))

Note that we *could* use the chain rule; however, we would then be dealing with three functions: ln, sin, and ^2.
How about we use log properties instead?

y = ln( [sin(t)]^2 )

As per the log property, we can move the exponent within the log to the outside.

y = 2 ln(sin(t))

Now we can differentiate, noting that we can leave the constant 2 alone. We still use the chain rule, but we only use it two times instead of three.

dy/dx = 2 [ 1/sin(t)] cos(t)

Simplify even further,

dy/dx = 2 [cos(t)/sin(t)]

Which, by definition, is equal to

dy/dx = 2cot(t)

You use the chain rule, because you have 2 inner functions within the outer function. That means we'll have 3 derivatives multiplied by one another.

d/dy = (1/(sin^2(theta)) * (2sin(theta)) * (cos(theta))

You can simplify it or whatever to make it look pretty. But what I did there was from left to right, I have the outermost, then innermost functions. The derivative of a ln(x) function is 1/x. The derivative of a x^2 function is 2x. Finally, the derivative of a sin(x) function is cos(x).

y = ln(sin^2(t)) = 2*ln(sin(t))

dy/dt = 2*cot(t)

y = ln [ (sin Ө)² ]

dy/dӨ = 2 sinӨ cos Ө / [ (sin Ө)² ] = 2 cot Ө