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

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Answer 1 · 2015-05-13T12:19:12

In order to derivate a #ln# we must remember its derivation rule: #dlnf(x) = (f'(x))/f(x)#

Let's just rewrite #sqrt(x)# as #x^(1/2)#. Like this: #ln(x^(1/2))#

Now, let's derivate it.

#f'(x) = (1/2).x^(-1/2)#

#f'(x) = (1/(2x^(1/2)))#

Now we've found #f'(x)#, let's solve the first derivative:

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

Simplifying it for exponential rules: #dln(x^(1/2)) = 1/(2x)#

Deriving #(1/(2x))#, again, to obtain your original function's second derivative:

#(d²ln(x^(1/2)))/(dx²) = (-1/(2x^(2)))# (or, alternatively, #-1/2 x^(-2)#)

Answer 2 · 2015-05-13T14:51:26

#sqrtx = x^(1/2)#, so

#ln (sqrtx)=ln (x^(1/2))=1/2lnx#

So the derivative is #1/2*1/x=1/2 x^(-1)#

The second derivative, then, is

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