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

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Answer 1 · 2017-11-26T06:52:54

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

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. $(dy)/(dx)=1/(1+(1-x)/(1+x))xx(-sqrt((1+x)^3)/sqrt(1-x))$

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

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

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