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