#d/dx log_x x#?

1 Answer
Aug 31, 2017

#0 " (forĀ "x>0).#

Explanation:

#d/(dx)(log_10(10))=d/(dx)(1)=0#

#y=log_x x" "<=>" "x^y=x#

#:.y=1=>log_x x=1#, for #x>0#, since #log_b x# functions are undefined for #x<=0# and #b<=0#.

#:.d/(dx)(log_x x)=0," "x>0#