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Picknell14
Picknell14 Picknell14
  • 22-04-2015
  • Mathematics
Answered

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

Verified Answer :

kittymaniac
kittymaniac kittymaniac
  • 22-04-2015

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

\(y = ln(sec(x)+tan(x)) \\ \\ y' = \frac{1}{sec(x)+tan(x)} * [tan(x)sec(x) + sec^2(x)] \\ \\ y' = \frac{sec(x)(sec(x) + tan(x))}{sec(x)+tan(x)} \\ \\ y' = sec(x) + C\)

I hope this helps! :)

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