Question

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

Answer 1

Call the function `f(x)`. By laws of logarithms and the definition `ln(e) = 1`, we have

`f(x) = x/(e^x - 1)`

We now use the quotient rule to find the derivative. If `f(x) = g(x)/(h(x))`, then `f'(x)= (g'(x)h(x) - h'(x)g(x))/(h(x))^2`.

Here we have `g(x) = x` and `h(x) = e^x - 1`. Then `g'(x) = 1` and `h'(x) = e^x`.

`f'(x) = (x(e^x - 1) - e^x(x))/(e^x - 1)^2`

`f'(x) = (xe^x - x - xe^x)/(e^x - 1)^2`

`f'(x) = -x/(e^x - 1)^2`

Hopefully this helps!