`" If "y=sqrt((1-x)/(1+x))," then "(1=x^(2))(dy)/(dx)` is equal to

Options

A. `y^(2)`
B. `1//y`
C. `-y`
D. `-y//x`

Explanation

`"We have "y=sqrt((1-x)/(1+x)`
Differentiating w.r.t. x, we get
`(dy)/(dx)=(1)/(2)((1-x)/(1+x))^(1//2-1)(d)/(dx)((1-x)/(1+x))`
`=(1)/(2)sqrt((1+x)/(1-x))cdot((1+x)(-1)-(1-x)(1))/((1+x)^(2))`
`=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))(1-x)^(2)`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1-x)/(1+x))`
`"or "(1-x^(2))(dy)/(dx)=-y`
`"or "(1-x^(2))(dy)/(dx)+y=0`