## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 437: 57

#### Answer

$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$

#### Work Step by Step

Differentiate $f(x)=ln(x-1)$ with respect to $x$. $f'(x)=\frac{1}{(x-1)}$ To find the nth differentiation, we will take possible derivative for getting general formula such as: $f''(x)=(-1)(x-1)^{-2}$ $f'''(x)=(-1)^{2}1.2(x-1)^{-3}$ After solving in the same manner, we find that $f^{n}(x)=(-1)^{n-1}1.2.3....(n-1)(x-1)^{-n}$ Hence,$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$

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