Whats the derivative of $y = \frac{1}{2} ln [\frac{1 + x}{1 - x}]$ $y=1/2 ln[(1+x)/(1-x)]$ ?

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Answer 1 · 2017-09-03T11:38:26

$\frac{1}{1 - x^{2}} .$ $1/(1-x^2).$

$y = \frac{1}{2} ln [\frac{1 + x}{1 - x}] .$ $y=1/2ln[(1+x)/(1-x)].$

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

$y = \frac{1}{2} [ln (1 + x) - ln (1 - x)] .$ $y=1/2[ln(1+x)-ln(1-x)].$

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

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

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

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

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

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

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

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