Find a formula for fnx if fxlnx1Solutionfx lnx 1 fx 1x 1

Find a formula for f^(n)(x) if f(x)=ln(x-1)

Solution

f(x) = ln(x - 1)
f\'(x) = 1/(x - 1) = (x - 1)^(-1)
f\'\'(x) = (-1)(x - 1)^(-2)
f\'\'\'(x) = (-1)(-2)(x - 1)^(-3)
f\'\'\'\'(x) = (-1)(-2)(-3)(x - 1)^(-4).

From here, we can see that each successive differentiation gives a coefficient of (-1)^(n - 1) * (n - 1)! and then the power of x - 1 is -n. Thus:

f^(n)(x) = (-1)^(n - 1) * (n - 1)! * (x - 1)^(-n).

To prove this, we need to show that if f^(n)(x) = (-1)^(n - 1) * (n - 1)! * (x - 1)^(-n), then:

f^(n+1)(x) = (-1)^n * n! * (x - 1)^(-n - 1).

Differentiating f^(n)(x) gives:

d/dx f^(n)(x) = f^(n+1)(x)
= d/dx [(-1)^(n - 1) * (n - 1)! * (x - 1)^(-n)]
= (-1)^(n - 1) * (n - 1) * (-n)(x - 1)^(-n - 1)
= -1 * (-1)^(n - 1) * n(n - 1)! * (x - 1)^(-n - 1)
= (-1)^n * n! * (x - 1)^(-n - 1), which is what we wanted to prove.