`y=ln|secx+tanx|`

To take the derivative of this function, use the formula:

`(ln u)' = 1/u* u'`

Applying this formula, y' will be:

`y' = 1/(secx+tanx) * (secx+tanx)'`

To get the derivative of the inner function, use the formulas:

`(sec theta)'= sec theta tan theta`

`(tan theta)' =sec^2 theta`

So y' will become:

`y' = 1/(secx +tanx) * (secxtanx+sec^2x)`

Simplifying it, the derivative will be:

`y'=(secxtanx+sec^2x)/(secx+tanx)`

`y'=(secx(tanx+secx))/(secx + tanx)`

`y'=(secx(secx+tanx))/(secx+tanx)`

`y'=secx`

**Therefore, the derivative of the given function is `y' =secx` .**

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