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How do you differentiate given #y = (sec)^2 x + (tan)^2 x#?

1 Answer

Mar 2, 2018

#y' = 4sec^2xtanx#

Explanation:

#y = sec^2x + tan^2x#
#y = secx secx + tanx tanx#

#y' = (dy)/(dx) = (secxtanx)secx+secx(secxtanx)+ (sec^2x)tanx+tanx(sec^2x)#

#y' = sec^2xtanx + sec^2xtanx+sec^2xtanx+sec^2xtanx#

#y' = 4sec^2xtanx#

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