1
$\begingroup$

I got $y'= x^2 + \ln({1\over x})\times ({1\over x})^x$

Am I correct?

So $y'=-x^{-x}(\ln{x} + 1)$

  • 1
    Hint: Take log on both sides.2012-10-17
  • 0
    You can check your work here: http://www.wolframalpha.com/input/?i=derivative+of+1%2Fx%5Ex2012-10-17
  • 0
    why can't i use the natural log?2012-10-17
  • 0
    @ data: You $should$ use natural log.2012-10-17

2 Answers 2