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

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Answer 1 · 2015-05-14T12:37:14

Remembering the rule to derive #ln# functions:

If #y = ln(f(x))#, #(dy)/(dx) = (f'(x))/(f(x))#.

And using the chain rule, we can say that #u=x^(1/2)#

Now, #(dy)/(du) = (u')/(u)#

#u' = (1/2)*x^(-1/2) = 1/(2x^(1/2))#

Substituting in our original derivative:

#(dy)/(dx) = (1/(2x^(1/2)))/x^(1/2)#

#(dy)/(dx) = 1/(2x^(1/2))*1/x^(1/2) = 1/(2x)#

Answer 2 · 2015-05-14T13:12:18

There are a couple of methods available. Here is one:

Use properties of #ln# to rewrite:

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

So the derivative is

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