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)
