Solve D/dx (ln ( 1/cos(x) + tan (x) )

Solve D/dx (ln ( 1/cos(x) + tan (x) ) As always, we approach with a substitution method that we would normally use for differentiating ln (x). So, we try differentiate ln (t) with t= 1/cos(x) + tan(x)So we, would have to differentiate t as the differentiated form of ln(t) is 1/t * dy/dtSo, taking t= 1/cos(x) + tan (x), we do each part individually, and end up with dy/dt= ((cos(x))^-2*)sin(x) (via normal differentiating rules) + sec^2(x) (as tan(x) differentiates to (sec^2) Putting in the form 1/t * dy/dt,We get (Cos(x)^-2)*sin(x) +sec^2(x)) / (1/cos(x) + tan(x))The rest is simplification!the above = (sin(x)/ cos^2(x) + sec^2(x)) / (1/cos(x) + tan(x))(As sin/cos becomes tan multiplied with an extra 1/cos which becomes sec (first term of the numerator))= (tan(x)sec(x) + sec^2(x)) / (sec(x) + tan(x))(by factorising out sec(x)at the numerator)= (sec(x)) (tan(x) + sec(x)) / (tan(x) +sec(x)) Top and bottom cancel out to become final answer= sec(x)!