Question

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

Answer 1

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))`

Answer 2

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)`