How do you find the derivative of #y=ln(sqrt(x))#?
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2 Answers
Aug 18, 2014
#y'=1/(2x)# Explanation
#y=ln(sqrtx)# differentiating with respect to
#x# using Chain Rule,
#y'=1/sqrt(x)*1/2x^(-1/2)#
#y'=1/sqrt(x)*(1/(2sqrt(x)))#
#y'=1/(2x)#
Nov 6, 2015
