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

1 Answer
Jul 31, 2015

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#