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

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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.

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