`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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