How do I find the derivative of #y=ln(secx + tanx)#?

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How do I find the derivative of #y=ln(secx + tanx)#?

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Answer 1 · 2015-02-01T15:45:48

You may use the Chain Rule; you start deriving the logarithm as a normal one and then multipy by the derivative of its argument (which basically is #1/cos(x)+sin(x)/cos(x)#):
#y'=1/(sec(x)+tan(x))*[sin(x)/cos^2(x)+(cos^2(x)+sin^2(x))/cos^2(x)]=#

#=1/(sec(x)+tan(x))*[1+sin(x))/(cos^2(x))=1/cos(x)#

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How do I find the derivative of #y=ln(secx + tanx)#?

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