# How do you find the derivative of #ln((x+1)/(x-1))#?

I think it's simplest to rewrite first.

So

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To find the derivative of ln((x+1)/(x-1)), you can use the chain rule and the quotient rule. Here's the step-by-step process:

- Let y = ln((x+1)/(x-1)).
- Rewrite the expression using properties of logarithms: y = ln(x+1) - ln(x-1).
- Differentiate each term separately using the chain rule:
- For ln(x+1), the derivative is (1/(x+1)) * (d/dx)(x+1).
- For ln(x-1), the derivative is (1/(x-1)) * (d/dx)(x-1).

- Apply the quotient rule for the second term: (f'(x)g(x) - f(x)g'(x)) / [g(x)]^2.
- Substitute the derivatives and simplify.

The final derivative is: [ \frac{d}{dx} \ln\left(\frac{x+1}{x-1}\right) = \frac{1}{x+1} - \frac{1}{x-1} ]

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