Question

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

Answer 1

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

Explanation:

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.