Find the derivative
$y=({1\over x})^x$
1
$\begingroup$
I got
$y'= x^2 + \ln({1\over x})\times ({1\over x})^x$
Am I correct?
So
$y'=-x^{-x}(\ln{x} + 1)$
implicit-differentiation
asked
2012-10-17
user id:41766
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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%5Ex
–
2012-10-17
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
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