Find `(dy)/(dx)` for `y=tan^(-1)sqrt((a-x)/(a+x))` `,-a`<`x`<`a`

Explanation

`y=tan^(-1){sqrt((a-x)/(a+x))}," where " -a lt x lt a`
Substituting `x= a cos theta,` we get
`y=tan^(-1){sqrt((a-a cos theta)/(a+a cos theta))},`
`=tan^(-1){sqrt((1- cos theta)/(1+ cos theta))},`
`=tan^(-1){sqrt(tan^(2)""(theta)/(2))}`
`=tan^(-1)|tan""(theta)/(2)|`
Also, for `-a lt x lt a, -1 lt cos theta lt 1`
`"or "theta in (0,pi) or (theta)/(2) in (0, (pi)/(2))`
`therefore" "y=tan^(-1)|tan""(theta)/(2)|=tan^(-1)(tan""(theta)/(2))`
`=(theta)/(2)=(1)/(2) cos^(-1) ((x)/(a))`
`"or "(dy)/(dx)=-(1)/(2)xx(1)/(sqrt(1-(x^(2))/(a^(2))))(d)/(dx)((x)/(a))=-(1)/(2sqrt(a^(2)-x^(2)))`