How do I differentiate #y=ln(sec(x) tan(x))#?

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How do I differentiate #y=ln(sec(x) tan(x))#?

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Answer 1 · 2015-02-22T01:06:19

You use the Chain Rule deriving first #ln# as it is and then multiplying by the derivative of the argument and also the Product Rule where:
if #f(x)=g(x)*h(x)#
#f'(x)=g'(x)h(x)+g(x)h'(x)#

You get:

#y'=1/(sec(x)tan(x))*[sec(x)tan(x)tan(x)+sec(x)sec^2(x)]=#

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

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How do I differentiate #y=ln(sec(x) tan(x))#?

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