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How do you find the derivative of #ln(sqrtx)#?

1 Answer

Dec 19, 2015

#1/(2x)#

Explanation:

The chain rule can be generalized to terms in the natural logarithm as follows:

#d/dx[ln(u)]=(u')/u#

Thus,

#ln(sqrtx)=ln(x^(1/2))=(d/dx[x^(1/2)])/x^(1/2)#

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

#=1/(2x)#

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