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

Question

Elijah Abram · Accepted Answer

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!

Ezra Adams · Answer

undefined 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} $.