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

Question

Emily Ammerman · Accepted Answer

Simplify using natural log properties, take the derivative, and add some fractions to get #d/dxln((x+1)/(x-1))=-2/(x^2-1)# It helps to use natural log properties to simplify #ln((x+1)/(x-1))# into something a little less complicated. We can use the property #ln(a/b)=lna-lnb# to change this expression to: #ln(x+1)-ln(x-1)# Taking the derivative of this will be a lot easier now. The sum rule says we can break this up into two parts: #d/dxln(x+1)-d/dxln(x-1)# We know the derivative of #lnx=1/x#, so the derivative of #ln(x+1)=1/(x+1)# and the derivative of #ln(x-1)=1/(x-1)#: #d/dxln(x+1)-d/dxln(x-1)=1/(x+1)-1/(x-1)# Subtracting the fractions yields: #(x-1)/((x+1)(x-1))-(x+1)/((x-1)(x+1))# #=((x-1)-(x+1))/(x^2-1)# #=(x-1-x-1)/(x^2-1)# #=-2/(x^2-1)#

Paisley Johnson · Answer

undefined To find the derivative of $ \ln\left(\frac{x+1}{x-1}\right) $, you can use the chain rule and the properties of logarithmic functions. The derivative is given by:

$$ \frac{d}{dx} \left[ \ln\left(\frac{x+1}{x-1}\right) \right] = \frac{1}{\frac{x+1}{x-1}} \cdot \frac{d}{dx} \left[ \frac{x+1}{x-1} \right] $$

Using the quotient rule to differentiate $ \frac{x+1}{x-1} $, we get:

$$ \frac{d}{dx} \left[ \frac{x+1}{x-1} \right] = \frac{(x-1) \cdot (1) - (x+1) \cdot (1)}{(x-1)^2} $$

Simplify the expression:

$$ \frac{d}{dx} \left[ \frac{x+1}{x-1} \right] = \frac{x-1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2} $$

Substitute this back into the derivative of $ \ln\left(\frac{x+1}{x-1}\right) $:

$$ \frac{d}{dx} \left[ \ln\left(\frac{x+1}{x-1}\right) \right] = \frac{1}{\frac{x+1}{x-1}} \cdot \frac{-2}{(x-1)^2} $$

$$ = \frac{-2(x-1)}{(x+1)} \cdot \frac{(x-1)}{(x-1)^2} $$

$$ = \frac{-2}{x+1} $$

So, the derivative of $ \ln\left(\frac{x+1}{x-1}\right) $ is $ \frac{-2}{x+1} $.