Short Answer: It's a lot like showing #d/(dx)(cosx)=-sinx#
Medium Answer:
Use #d/(dx)(f(x))=lim_(hrarr0)(f(x+h)-f(x))/h#, and
#sect=1/cost# and
#cos(x+h)=cosx cosh - sinx sinh#
and the limits:
#lim_(hrarr0)sinh/h=1# and #lim_(hrarr0)(1-cosh)/h=0#.
Long Answer
#d/(dx)(secx)=d/(dx)(1/cosx) = lim_(hrarr0)(1/(cos(x+h))-1/cos(x))/h#
# = lim_(hrarr0)((cosx-cos(x+h))/(hcos(x)cos(x+h)))#
# = lim_(hrarr0)(cosx-(cosxcosh-sinxsinh))/(hcos(x)cos(x+h))#
# = lim_(hrarr0)[(cosx-cosxcosh)/(hcosxcos(x+h))+(sinxsinh)/(hcos(x)cos(x+h))]#
# = lim_(hrarr0)[cosx/(cosxcos(x+h))((1-cosh)/h)+(sinx)/(cos(x)cos(x+h))(sinh/h)]#
# = cosx/(cosxcos(x))(0)+(sinx)/(cos(x)cos(x))(1)#
#=sinx/cos^2x=1/cosxsinx/cosx=secxtanx#
