We have #d/dx1/2ln(x)#, which we can write as:

#1/2d/dxln(x)#

As #d/dxln(x)=1/x#, we can write:

#1/2*1/x#

#1/(2x)#

We have our derivative.

Maybe you want to know **why** #d/dxln(x)=1/x#.

Remember that the basic definition of a derivative is:

#lim_(xirarr0)(f(x+xi)-f(x))/xi#

Here, #f(x)=ln(x)#.

We have:

#lim_(xirarr0)(ln(x+xi)-ln(x))/xi#

Let's leave the limit out for now.

#(ln(x+xi)-ln(x))/xi#

#1/xiln((x+xi)/x)#

#1/xiln(1+xi/x)#

#ln(1+xi/x)^(1/xi)#

Taking #alpha=xi/x#, and so #xi=alphax#, we can rewrite our limit calculation as:

#lim_(alphararr0)ln(1+alpha)^(1/(alphax))#

#lim_(alphararr0)1/xln(1+alpha)^(1/alpha)#

Remember that #e=lim_(nrarroo)(1+1/n)^n#. Another definition of the above is:

#lim_(nrarr0)(1+n)^(1/n)#, which is what we have here. We write the above as:

#1/x*ln(e)#, and as #ln(e)=1#, this reduces to:

#1/x#, the derivative of #ln(x)#