How do you differentiate #y=ln(secx tanx)#?

1 Answer
Mar 15, 2015

Use the derivative of #lnx# and the Chain Rule.

#d/(dx)(lnx)=1/x#, so (by the chain rule)

#d/(dx)(ln(u))=1/u*(du)/(dx)#

#y=ln(secxtanx)#

#y'=1/(secxtanx)*d/(dx)(secxtanx)#

#=1/(secxtanx)*((secxtanx)tanx+secx*sec^2x)#

(I use the product rule in the form: #d/(dx)(FS)=F'S+FS'#)

#y'=1/(secxtanx)(secxtan^2x+sec^3x)#

There are many other ways to write #1/(secxtanx)(secxtan^2x+sec^3x)#. One of the more obvious is:

#(secxtan^2x+sec^3x)/(secxtanx)=(tan^2x+sec^2x)/tanx# which

could also be written: #tanx+sec^2x/tanx=tanx+secxcscx#. or, factoring out #1/cosx#, we could write #secx(sinx+cscx)# and so on.
(And trig students ask why we have to do identities).