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

User Jamie Cockburn
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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! :)
User Prayag Choraria
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