Given f(x) = (x-3) / ln x

We need to find the derivative of f(x).

We notice that f(x) is a quotient of two functions.

Then, we will use the quotient rule to find f'(x).

==> Let f(x) = u/v such that:

u= x-3 ==> u' = 1

v = lnx ==> v' = 1/x

Then we know that:

f'(x) = (u'v - uv')/v^2

= ( 1*lnx - (x-3)*1/x) / (lnx)^2

= (lnx - 1 + 3/x ) /(lnx)^2

**==> f'(x) = (xlnx -x + 3)/x(lnx)^2**

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}.