For #y=ln(sec x + tan x)#, what is #y'# and #y''#?

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Answer 1 · 2018-03-05T23:00:44

#dy/dx =secx#

#(d^2y)/dx^2 = secxtanx#

#y=ln(secx+tanx)#

using the chain rule:

#dy/dx = 1/(secx+tanx) d/dx (secx+tanx)#

#dy/dx = 1/(secx+tanx) (secx tanx+sec^2x)#

#dy/dx = (secx (secx +tanx))/(secx+tanx) = secx#

and then:

#(d^2y)/dx^2 = secxtanx#