Help me find y' and y''. y= ln(sec 6x + tan 6x)

y = ln ( sec 6x + tan 6x)

y'=

y''=

Thank you!

Recall that d/dx[ln(u)] = 1/u * (du/dx)

y = ln(sec(6x) + tan(6x))

y' = [6*sec(6x)*tan(6x) + 6*sec²(6x)]/[sec(6x) + tan(6x)]

y' = (6*sec(6x))*(tan(6x) + sec(6x))/(sec(6x) + tan(6x))

The tan(6x) + sec(6x)'s cancel out.

y' = 6*sec(6x)

y'' = 6*sec(6x)*tan(6x) * 6 = 36*sec(6x)*tan(6x)

y' = 6 sec 6x

y" = 36 sec 6x tan 6x