Differentiate f(x)= (x-3)/(lnx)

To differentiate f(x) =
(x-3)/lnx.

f(x) =
(x-3)*(1/lnx).

We know that d/dx{u(x) (v(x)} = (d/dx) u(x)*
v(x)+u(x)(d/dx)v(x)....(1).

Here u(x) = x-3). (d/dx)u(x) =
(d/dx)(x-3) = 1.

v(x) = 1/lnx. (d/dx)v(x) = d/dx(1/lnx) =
(d/dx)(lnx)^(-1).

(d/dx)(lnx)^(-1) =
(-1)(lnx)^(-2)}{d/dx)(lnx).

(d/dx)(1/lnx) =
{-1/(lnx)^2}{1/x).

(d/dx)(1/lnx) =
-1/{x(lnx)^2}.

So we substitute the results u(x) = (x-3),
 (d/dx)u(x) = 1, v(x) = 1/lnx and (d/dx)(1/lnx) = -1/{x(lnx)^2} in (1) and
get:

(d/dx){(x-3)/lnx} = 1/lnx +
(x-3)[-1/{x(lnx)^2}].

(d/dx){(x-3)/lnx} = {xlnx -
(x-3)}/{x(lnx)^2}.

(d/dx){(x-3)/lnx} = (xlnx - x
+3)/{x(lnx)^2}.

Therefore (d/dx)(x-3)/(lnx) =
{x(lnx-1)+3}/{x(lnx)^2}.