Whats the derivative of #ln((1+x)/(1-x))#?

1 Answer
Nov 6, 2016

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

Explanation:

Let #y = lnu# and #u = (1 + x)/(1 - x)#.

We will use the chain rule to differentiate this problem. However, we must first find the derivative of each function.

#y' = 1/u#

By the quotient rule:

#u' = (1(1 - x) - (-1(1 + x)))/(1 - x)^2#

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

#u' = (1 - x + 1 + x)/(1- x)^2#

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

Now, by the chain rule:

#dy/dx= 1/u xx 2/(1 - x)^2#

#dy/dx= 1/((1 + x)/(1 - x)) xx 2/(1 - x)^2#

#dy/dx = (1 - x)/(1 + x) xx 2/(1 - x)^2#

#dy/dx= 2/((1 + x)(1 - x))#

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

Hopefully this helps!