Question

What is the derivative of `y=ln(cos^2ɵ)`?

Answer 1

Assuming that we want `y' = dy/(d theta)`, in simplest form: `y' = -2tan theta`

Explanation:

`y = ln(cos^2 theta)`

Method 1 Leave it as is and use the chain rule twice:

`y' = 1/cos^2 theta * d/(d theta) (cos^2 theta)`

`= 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta)`

`= 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta)`

`= 1/cos^2 theta * 2cos theta * (-sin theta)`

`= -2 sintheta/costheta = -2tan theta`

Method 2 Use properties of `ln` to rewrite:

`y = ln(cos^2 theta) = 2ln(cos theta)`

Use the chain rule: (less detail this time)

`y' = 2*1/cos theta *(-sin theta) = -2tan theta`

For derivative with respect to `x`

`dy/dx = dy/(d theta)* (d theta)/dx`

So

`dy/dx = -2tan theta (d theta)/dx`