How do you find the derivative of y= ln (x/(x-1))?

1 Answer
Feb 12, 2016

-1/(x^2-x)

Explanation:

The derivative of lnu, by the chain rule, is 1/u*u'. So we have to find the derivative of the inside function (which in our case is (x)/(x-1)) in order to find the derivative of the entire function. In math terms, we express this as:
dy/dx=1/((x)/(x-1))*(x/(x-1))'

dy/dx=(x-1)/x*(x/(x-1))'

Finding the derivative of x/(x-1) is going to be a little difficult because we need to use the quotient rule, which says:
d/dx(u/v)=(u'v-uv')/v^2, where (for our example) u=x and v=x-1.
Alright, let's get to work:
d/dxx/(x-1)=((x)'(x-1)-(x)(x-1)')/(x-1)^2
d/dxx/(x-1)=(x-1-x)/(x-1)^2
d/dxx/(x-1)=(-1)/(x-1)^2

Now we multiply this result by the derivative of ln(x/(x-1)), or (x-1)/x:
(x-1)/x*(-1)/(x-1)^2
Simplifying:
-1/(x(x-1))
Or, equivalently,
-1/(x^2-x)