What is the derivative of ln((x+1)/(x-1))?

2 Answers
May 14, 2015

In three moments: first, remember ln derivative rule. Second, chain rule. Third and finally, quocient rule.

Derivative rule: be y=ln(f(x)), (dy)/(dx) = (f'(x))/(f(x))

Now, let's consider u = ((x+1)/(x-1))

Deriving ln(u), we get, then, (u')/u.

But the derivative of u is the derivative of ((x+1)/(x-1)), which is calculated as follows:

(1*(x-1)-(x+1)*(1))/(x-1)^2 = (-2)/(x-1)^2

Remembering the values of u and u' and substituting them into the derivative of ln(u), we get:

(dy)/(dx) = ((-2)/(x-1)^2)/((x+1)/(x-1))

(dy)/(dx) = ((-2)*(x-1))/((x-1)^2 (x+1)) = (-2)/((x-1)(x+1))

May 14, 2015

There are several ways to get to the correct answer. Here is one:

Use properties of logarithm to rewrite:

y=ln ((x+1)/(x-1))=ln(x+1) - ln(x-1)

Now use d/dx(lnu)=1/u (du)/dx to get:

dy/dx = 1/(x+1) - 1/(x-1)

If you prefer to write the result as a single fraction, do so.

dy/dx = (-2)/(x^2-1)