How do you find the derivative of #lnx^(1/2)#?

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Answer 1 · 2016-09-08T15:19:28

#d/dxlnx^(1/2)=1/(2x)#

We can

either use the chain rule here

#d/dxlnx^(1/2)=d/dx^(1/2)(lnx^(1/2))xxd/dx x^(1/2)#

= #1/x^(1/2)xx1/2xx x^(-1/2)#

= #1/x^(1/2)xx1/2xx1/x^(1/2)=1/(2x)#

or as #lnx^(1/2)=1/2lnx#

#d/dxlnx^(1/2)=1/(2x)#