Question

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)`