What is the first (ln(x-1))^' ?

The first (ln(x-1))^' is 1/(x-1)

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(ln(x-1))^'

(ln (x - 1))^{^{'}}

\frac{1}{x - 1}

Solution steps

(ln (x - 1))^{'}

Apply the chain rule:

\frac{1}{x - 1} (x - 1)^{'}

= \frac{1}{x - 1} (x - 1)^{'}

(x - 1)^{'} = 1

= \frac{1}{x - 1} \cdot 1

Simplify

= \frac{1}{x - 1}

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