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

Answer 1

Christopher Allers

I think it's simplest to rewrite first.

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

Recall that #d/dx(lnu) = 1/u (du)/dx#, so

#d/dx(ln(x+1)) = 1/(x+1)# and #d/dx(ln(x-1)) = 1/(x-1)#

#dy/dx = [1/(x+1)-1/(x-1)]#

# = (-2)/(x^2-1)#

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Answer 2

Charles Anctil

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.