\( \dfrac{d}{dx}\ln\left(\dfrac{1}{1-x}\right)= \)

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Rewrite the expression. Bring the exponent to the front.

$$ \ln\frac{1}{1-x}$$ $$ \ln(1-x)^{-1} $$ $$ -\ln(1-x) $$

We can differentiate this expression using the chain rule.

$$ \frac{d}{dx}(-\ln(1-x)) =-\frac{1}{1-x}\cdot-1$$ $$ = \boxed{\frac{1}{1-x}}$$

Use the quotient rule of logarithms to rewrite the expression.

$$ \ln\frac{1}{1-x}$$ $$ =\ln1 - \ln(1-x)$$

Differentiate, using the chain rule for the second part.

$$ \frac{d}{dx}\ln1+\frac{d}{dx}(-\ln(1-x)) $$ $$ = 0 - \frac{1}{1-x} \cdot -1 $$ $$ = \boxed{\frac{1}{1-x}}$$