Question

What is the derivative of `y=arctan sqrt((1-x)/(1+x))`?

Answer 1

`(dy)/(dx)=-sqrt((1+x)^5)/(2sqrt(1-x))`

Explanation:

Let `u=sqrt((1-x)/(1+x))`. Also observe `(1-x)/(1+x)=1-(2x)/(1+x)`

then using chain rule `(du)/(dx)=1/(2sqrt((1-x)/(1+x)))xxd/(dx)(1-(2x)/(1+x))`

= `sqrt(1+x)/(2sqrt(1-x))xxd/(dx)(1-(2x)/(1+x))`

= `sqrt(1+x)/(2sqrt(1-x))xx(-(2(1+x)-2x)/(1+x)^2)`

= `sqrt(1+x)/(2sqrt(1-x))xx(-2/(1+x)^2)`

= `-sqrt(1+x)^3/sqrt(1-x)`

Hence `y=arctanu` and hence `(dy)/(dx)=1/(1+u^2)xx(du)/(dx)`

i.e. `(dy)/(dx)=1/(1+(1-x)/(1+x))xx(-sqrt((1+x)^3)/sqrt(1-x))`

= `-(1+x)/2xxsqrt((1+x)^3)/sqrt(1-x)`

= `-sqrt((1+x)^5)/(2sqrt(1-x))`