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.