What is the derivative of #y=ln(cos^2ɵ)#?
1 Answer
Jul 31, 2015
Assuming that we want
Explanation:
Method 1 Leave it as is and use the chain rule twice:
#= 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
Use the chain rule: (less detail this time)
For derivative with respect to
So