What is the derivative of #y=sec(2x) tan(2x)#?

1 Answer
WADU HEK
Apr 14, 2018

2sec(2x)#(sec^2(2x) + tan^2(2x))#

Explanation:

# y' = (sec(2x))(tan(2x))' + (tan(2x))(sec(2x))'# (Product Rule)
# y' = (sec(2x))(sec^2(2x))(2) + (tan(2x))(sec(2x)tan(2x))(2) # (Chain rule and derivatives of trig)
# y' = 2sec^3(2x) + 2sec(2x)tan^2(2x) #
# y' = 2sec(2x)(sec^2(2x) + tan^2(2x)) #

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