Just do the product rule, followed by the chain rule (on only the #ln(u)#). Recall that #d/(dx)[cosx] = -sinx#, and that #d/(dx)[ln(u(x))] = 1/(u(x))(du(x))/(dx) = 1/(u(x))((d)/(dx)[u(x)])#
So:
#d/(dx)[-cosx*ln(secx+tanx)]#
#= [-cosx*(1/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]#
Move the cosine in:
#= [(-cosx/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]#
Factor and cancel:
#= [(-cosx/cancel(secx+tanx))*{secx(cancel(tanx+secx))}]+[ln(secx+tanx)sinx]#
Notice that #secx = 1/cosx#:
#= [-cosx*secx]+[ln(secx+tanx)sinx]#
And you get:
#= -1+sinxln(secx+tanx)#