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

Answer 1

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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