How do you find the derivative of #y=1/2 ln[(1+x)/(1-x)]#?

1 Answer
Steve M
Nov 9, 2016

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

Explanation:

# y = 1/2ln((1+x)/(1-x)) #
# :. y = 1/2{ln(1+x)-ln(1-x)} # (law of logs)

Differentiating (using the chain rule) gives:

# dy/dx = 1/2{1/(1+x)-(-1)/(1-x)} #
# :. dy/dx = 1/2{ ( (1-x)+(1+x) )/((1+x)(1-x)) } #
# :. dy/dx = 1/2{ 2/(1-x^2) } #
# :. dy/dx = 1/(1-x^2) #

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