Question

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

Answer 1

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).