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!
