Differentiate f and find the domain of f

f(x) = x / (1 − ln(x − 9))

1) Domain: You cannot take the ln of a non-positive number, so x-9 > 0

Therefore x > 9.

In addition the denominator cannot be 0, but that would happen when ln(x-9) = 1

In that case x - 9 = e and x = 9 + e

So: {x | x ix real, x > 9, and x ≠ 9 + e } <--------- Domain

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Differentiation

Use the quotient rule: d(u/v)/dx = [(v*du/dx) - (u*dv/dx)] / (v^2)

u = x and du/dx = 1

v = (1 - ln(x-9)) ........ dv/dx = (-1 / (x-9)) * 1 = (1 / (9 - x)) (I multiplied numerator and denominator by -1.)

v^2 = 1 - (2*(ln(x-9))) + ((ln(x-9))^2)

Now just substitute this into the quotient rule:

((1 - ln(x-9)) * 1) - ((x * (1 / (9 - x)))
---------------------------------------… <------- Answer
1 - (2*(ln(x-9))) + ((ln(x-9))^2)

With (x-9) in numerators and denominators you can't simplify this.

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