How do you find the derivative of #x^(3/x)#?

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Answer 1 · 2016-08-08T22:40:03

#=3 x^(3/x - 2) (1-ln x)#

use logs cos that's ugly!

#y = x^(3/x)#

#ln y = ln x^(3/x) = 3/x ln x#

then implicit diff.... plus product rule on the RHS
#1/y \ y' = - 3/x^2 ln x + 3/x* 1/x#

#1/(x^(3/x)) \ y' = 3/x^2 (1-ln x)#

# y' =(x^(3/x)) 3/x^2 (1-ln x)#

#=3 (x^(3/x)) x^(-2) (1-ln x)#

#=3 x^(3/x - 2) (1-ln x)#