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

1 Answer
Apr 12, 2018

f'(x)=(1-ln(x-1)+1/(x-1))/(1-ln(x-1))^2

Explanation:

Use the quotient rule

(d(g(x))/(h(x)))/(dx)=(g'(x)h(x)-g(x)h'(x))/(h(x))^2

Here

g(x)=x

g'(x)=1

h(x)=1-ln(x-1), and

h'(x)=-1/(x-1)

So

f'(x)=(1-ln(x-1)-(1)(-1/(x-1)))/(1-ln(x-1))^2

f'(x)=(1-ln(x-1)+1/(x-1))/(1-ln(x-1))^2