How do you find the derivative of #lnsqrt x#?

Question

1458 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-09T20:53:08

In general
#(d ln(a))/(da) = 1/a#

Specifically if #a=sqrt(x)#
#(d ln(sqrt(x)))/(d sqrt(x)) = 1/sqrt(x)#

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

#(d sqrt(x))/dx = 1/2 x^(-1/2) = 1/(2sqrt(x))#

#(d ln(sqrt(x)))/(dx) = (d ln(sqrt(x)))/(d sqrt(x)) * (d sqrt(x))/dx#

#= 1/sqrt(x) * 1/(2sqrt(x))#

#= 1/(2x)#

That is, the derivative of #ln sqrt(x)#
is
#1/(2x)#