Whats the derivative of #y=1/2 ln[(1+x)/(1-x)]#?

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Answer 1 · 2017-09-03T11:38:26

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

#y=1/2ln[(1+x)/(1-x)].#

Using the Familiar Rules of #log.# fun., we get,

#y=1/2[ln(1+x)-ln(1-x)].#

#:. dy/dx=1/2[d/dx{ln(1+x)}-d/dx{ln(1-x)}],#

#=1/2[1/(1+x)*d/dx(1+x)-1/(1-x)*d/dx(1-x)]...[because," the Chain Rule"],#

#=1/2[1/(1+x)(1)-1/(1-x)(-1)],#

#=1/2[1/(1+x)+1/(1-x)],#

#=1/2[{(1+x)+(1-x)}/{(1+x)(1-x)}],#

#=1/cancel2[cancel2/(1-x^2)].#

#:. dy/dx=1/(1-x^2),# is the desired Derivative.

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