What is the derivative of #f(x) = x/(1-ln(x-1))#?

Question

Elijah Abram · Accepted Answer

#f'(x)=(2x-1-(x-1)ln(x-1))/((x-1)[(1-ln(x-1)]^2# The complete solution is here #f(x)=x/(1-ln(x-1)# #f'(x)=(1(1-ln(x-1))-x(0-1/(x-1)))/([(1-ln(x-1)]^2# #f'(x)=((x-1)[1-ln(x-1)]+x)/((x-1)[(1-ln(x-1)]^2# #f'(x)=(x-1-(x-1)ln(x-1)+x)/((x-1)[(1-ln(x-1)]^2# #f'(x)=(2x-1-(x-1)ln(x-1))/((x-1)[(1-ln(x-1)]^2#

Isabella Ackerman · Answer

undefined To find the derivative of $ f(x) = \frac{x}{1 - \ln(x - 1)} $, we can use the quotient rule. The quotient rule states that if we have a function $ u(x) $ divided by another function $ v(x) $, then the derivative is given by $ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $.

Using the quotient rule:

Let $ u(x) = x $ and $ v(x) = 1 - \ln(x - 1) $.

Then, $ u'(x) = 1 $ and $ v'(x) = \frac{1}{x - 1} $ (since the derivative of $ \ln(x - 1) $ is $ \frac{1}{x - 1} $).

Now, applying the quotient rule:

$ 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} $

This expression represents the derivative of the given function $ f(x) = \frac{x}{1 - \ln(x - 1)} $.