How do you find the derivative of $arctan sqrt [ (1-x)/(1+x)]$ ?

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Answer 1 · 2016-08-09T10:06:33

$\ y' = - 1/(2 sqrt [(1-x^2)]$

$y = arctan sqrt [ (1-x)/(1+x)]$

$tan y = sqrt [ (1-x)/(1+x)]$

$tan^2 y = (1-x)/(1+x)$

$tan^2 y = 2/(1+x) - 1$

differentiating each side

$2 tan y sec^2 y\ y' = -2/(1+x)^2$

$tan y (tan^2 y + 1)\ y' = -1/(1+x)^2$

$sqrt [ (1-x)/(1+x)] ( (1-x)/(1+x) + 1)\ y' = -1/(1+x)^2$

$sqrt [ (1-x)/(1+x)] ( (2)/(1+x))\ y' = -1/(1+x)^2$

$\ y' = - 1/(1+x)^2 * sqrt [ (1+x)/(1-x)] * (1+x)/2$

$\ y' = - 1/(2 sqrt(1+x) * sqrt [(1-x)]$

$\ y' = - 1/(2 sqrt (1-x^2)$

How do you find the derivative of arctan sqrt [ (1-x)/(1+x)]arctan sqrt [ (1-x)/(1+x)]?