I WROTE THE QUESTION ON THE TITLE SECTION BECAUSE IT WASN\'T POSTING CLEARLY!PLE

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u = x, v = 1-ln(x-1), so v' = -1/(x - 1) and u' = 1 Derivative of quotient (u/v)' = [vu' -uv']/v^2x/(1-ln(x-1))? (u/v)' = {[1-ln(x-1)]*1 - x*-1/(x - 1)}/[1 - ln(x-1)]^2 multiply top and bottom by (x - 1) (u/v)' = {[1-ln(x-1)]*1 - x*-1/(x - 1)} = [x - 1 - (x-1) ln(x-1) + x]/{(x - 1)[1 - ln(x-1)]^2} (u/v)' = [2x - 1 - (x-1) ln(x-1)]/{(x-1)[1 - ln(x-1)]^2} [Background notes: Essentially the domain of some f(x) refers to the x values it may exist in. For instance, the domain of sine is the set of all real numbers, R, whereas the domain of the square root (x) consists only of numbers >= 0 Sometimes we extend or "tweak" a function, redefining it to allow it to exist for all real numbers. For example reciprocal (x), Extending the definition of f to f(x) = 1/x, for x ? 0, f(0) = 0, then f is defined for all real numbers, and its domain is R.] Now f(x) = x/(1-ln(x-1)) is only real for positive x values Also it has a similar problem to 1/x when denominator is zero i.e. when ln(x-1) = 1 or x - 1 = e^1 or x = e + 1 ~ 3.71828 So copying above, by redefining f(x) = 1/x, for x ? e + 1 f(e + 1) = 0 we may, (only then), allow the domain of f to be defined for all positive real numbers. (N.B. This is sometimes expressed in a more sophisticated way using the jargon of sets)

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