`xsqrt(1+y)+ysqrt(1+x)=0` then `(dy)/(dx)=`A. `-(1)/(1+x)`B. `-(1)/((1+x)^(2))`C. `(1)/((1+x)^(2))`D. `(sqrtx)/(sqrt(1+x))`

`xsqrt(1+y)+ysqrt(1+x)=0` then `(dy)/(dx)=`A. `-(1)/(1+x)`B. `-(1)/((1

1.	`xsqrt(1+y)+ysqrt(1+x)=0` then `(dy)/(dx)=`A. `-(1)/(1+x)`B. `-(1)/((1+x)^(2))`C. `(1)/((1+x)^(2))`D. `(sqrtx)/(sqrt(1+x))`
Answer» Correct Answer - B Given equation `xsqrt(1+y)+ysqrt(1+x)=0` Can be written as : `xsqrt(1+y)=-ysqrt(1+x)` Squaring both sides, we get `x^(2)(1+y)=y^(2)(1+x)` `rArr" "x^(2)+x^(2)y=y^(2)+y^(2)x" "rArr" "x^(2)-y^(2)=y^(2)x-x^(2)y` `rArr" "(x-y)(x+y)=-xy(x-y)` `rArr" "x+y=-xy" "rArr" "y(1+x)=-x` `y=(-x)/(1+x)` which is in explicit form. Differentiating w.r.t. x, we get `(dy)/(dx)=((1+x)(-1)+x(1))/((1+x)^(2))=(-1)/((1+x)^(2))`