How do you differentiate #ln((x+1)/(x-1)) #?

Answer 1

#d/dxln((x+1)/(x-1))=1/(x+1)-1/(x-1)#

To avoid needing to use the quotient rule , we will use the property of logarithms that #log(a/b) = log(a)-log(b)#
After that, we will use the chain rule as well as the known derivative #d/dxln(x) = 1/x#
#d/dxln((x+1)/(x-1)) = d/dx(ln(x+1)-ln(x-1))#
#=d/dxln(x+1)-d/dxln(x-1)#
#=1/(x+1)(d/dx(x+1))-1/(x-1)(d/dx(x-1))#
#=1/(x+1)(1)-1/(x-1)(1)#
#=1/(x+1)-1/(x-1)#
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Answer 2

To differentiate ( \ln\left(\frac{x+1}{x-1}\right) ), you can use the chain rule and the quotient rule. The derivative is:

[ \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] ]

Apply the quotient rule to find ( \frac{d}{dx}\left[\frac{x+1}{x-1}\right] ):

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

[ = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} ]

Now, simplify and combine terms:

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

[ = \frac{-2}{(x-1)^2} ]

Putting it all together:

[ \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)} ]

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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.

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.

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.

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.

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