How do you find the derivative of $arctan [(1-x)/(1+x)]^(1/2)$ ?

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Answer 1 · 2016-07-22T06:52:55

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=-1/2*(sqrt(1-x^2))/((1-x^2))$

Formula for finding derivative of arctangent

$d/dx(tan^-1 u)=(1/(1+u^2))d/dx(u)$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=(1/(1+(sqrt((1-x)/(1+x)))^2))d/dx(sqrt((1-x)/(1+x)))$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$(1/(1+(1-x)/(1+x)))(1/(2(sqrt((1-x)/(1+x)))))*d/dx((1-x)/(1+x))$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$(1/(1+(1-x)/(1+x)))(1/(2(sqrt((1-x)/(1+x)))))*(((1+x)(-1)-(1-x)(1))/(1+x)^2)$

Simplify

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$((1+x)/(1+x+1-x))(1/2*sqrt((1+x)/(1-x)))*((-1-x-1+x)/(1+x)^2)$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$((1+x)/(2))(1/2*sqrt((1+x)/(1-x)))*((-2)/(1+x)^2)$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$(1/2*sqrt((1+x)/(1-x)))*((-1)/(1+x))$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=$
$(1/2*sqrt(((1+x)/(1-x))((1-x)/(1-x))))*((-1)/(1+x))$

$d/dx(tan^-1 sqrt((1-x)/(1+x)))=-1/2*(sqrt(1-x^2))/((1-x^2))$

God bless....I hope the explanation is useful.

How do you find the derivative of arctan [(1-x)/(1+x)]^(1/2)arctan [(1-x)/(1+x)]^(1/2)?