What is the derivative of #sec x#?

#sec(x)=1/cos(x)#

So we want to calculate

#d/dx1/cos(x)=d/dx(cos(x)^-1)#

for the chain rule this is equal to

#d/dx(cos(x)^-1)=-cos(x)^-2*d/dxcos(x)#

#=-1/cos(x)^2*(-sin(x))#

#=sin(x)/cos(x)^2#

or, if you prefer, it is

#=tan(x)sec(x)#.

To find the derivative of secant, we could either use the limit definition of the derivative (which would take a very long time) or the definition of secant itself:
#secx=1/cosx#

We know #d/dxcosx=-sinx# - keep that in mind because we're going to need it.

Our problem is:
#d/dxsecx#

Since #secx=1/cosx#, we can write this as:
#d/dx1/cosx#

We can find this derivative using the quotient rule:
#d/dxu/v=(u'v-uv')/v^2#

In our case, #u=1->u'=0# and #v=cosx->v'=-sinx#:
#d/dx1/cosx=((0)(cosx)-(1)(-sinx))/(cosx)^2#
#=sinx/cos^2x#

This is equivalent to:
#1/cosx*sinx/cosx#

Because #1/cosx=secx# and #sinx/cosx=tanx#, this is:
#secxtanx#

Therefore,
#d/dxsecx=d/dx1/cosx=secxtanx#

Pro Tip
You can use this method to find the derivative of #cscx# and #cotx# by recognizing #cscx=1/sinx# and #cotx=1/tanx# and using the quotient rule, as we did above. This way, you only have to memorize the derivatives of sine, cosine, and tangent - you can derive the other three.