# How do you differentiate #f(x) = x/(1-ln(x-1))#?

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To differentiate the function ( f(x) = \frac{x}{1-\ln(x-1)} ), you would use the quotient rule, which states that if ( f(x) = \frac{u(x)}{v(x)} ), then ( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ). Applying this rule:

( u(x) = x ) and ( v(x) = 1 - \ln(x-1) )

( u'(x) = 1 ) and ( v'(x) = -\frac{1}{x-1} )

( f'(x) = \frac{(1)(1-\ln(x-1)) - (x)(-\frac{1}{x-1})}{(1-\ln(x-1))^2} )

( f'(x) = \frac{1 - \ln(x-1) + \frac{x}{x-1}}{(1-\ln(x-1))^2} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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