Question

What is the derivative of y=ln(sec(x)+tan(x))y=ln(sec(x)+tan(x))?

Answer 1

The derivative of ln(u) is 1/u * the derivative of u.


`y = ln(sec(x)+tan(x)) \\ \\ y' = (1)/(sec(x)+tan(x)) * [tan(x)sec(x) + sec^2(x)] \\ \\ y' = (sec(x)(sec(x) + tan(x)))/(sec(x)+tan(x)) \\ \\ y' = sec(x) + C`

I hope this helps! :)