Start by rewriting #secx# and #tanx# in terms of sine and cosine. Call #ln|secx + tanx|, #f(x)#.
#f(x) = ln|secx + tanx|#
#f(x) = ln|1/cosx + sinx/cosx|#
#f(x) = ln|(1 + sinx)/cosx|#
Now use the rule #ln(a/b) = lna - lnb# to simplify further.
#f(x) = ln(1 + sinx) - ln(cosx)#
You can differentiate this using the chain rule on each term. I'll show you how to do one. Let #y = lnu# and #u = 1+ sinx#. Then #dy/(du) = 1/u# and #(du)/dx = cosx#. If #dy/dx = dy/(du) * (du)/dx#, then #dy/dx = 1/u * cosx = cosx/(1 + sinx)#.
#f'(x) = cosx/(1 + sinx) - (-sinx/cosx)#
#f'(x) = cosx/(1 + sinx) + sinx/cosx#
This can be simplified.
#f'(x) = (cosx(cosx))/(cosx(1 + sinx)) + (sinx(1 + sinx))/(cosx(1 +sinx))#
#f'(x) = (cos^2x + sinx + sin^2x)/(cosx(1 + sinx))#
Apply #sin^2x + cos^2x = 1#.
#f'(x) = (1 + sinx)/(cosx(1 + sinx))#
#f'(x) = 1/cosx#
Apply #secx = 1/cosx#.
#f'(x) = secx#
Hopefully this helps!
