# How do you find the derivative of #(ln * e^x)/(e^x-1)#?

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To find the derivative of ( \frac{\ln(e^x)}{e^x - 1} ), you can use the quotient rule of differentiation. The quotient rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their quotient, ( \frac{u}{v} ), is given by ( \frac{u'v - uv'}{v^2} ).

First, let's find the derivatives of the numerator and denominator:

For the numerator ( \ln(e^x) ), we have: [ \frac{d}{dx}(\ln(e^x)) = \frac{d}{dx}(x) = 1 ]

For the denominator ( e^x - 1 ), we have: [ \frac{d}{dx}(e^x - 1) = e^x ]

Now, applying the quotient rule, we have: [ \frac{d}{dx}\left(\frac{\ln(e^x)}{e^x - 1}\right) = \frac{(1)(e^x - 1) - (\ln(e^x))(e^x)}{(e^x - 1)^2} ]

Expanding and simplifying, we get: [ \frac{e^x - e^x + \ln(e^x)e^x}{(e^x - 1)^2} ] [ = \frac{\ln(e^x)e^x}{(e^x - 1)^2} ]

So, the derivative of ( \frac{\ln(e^x)}{e^x - 1} ) is ( \frac{\ln(e^x)e^x}{(e^x - 1)^2} ).

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