If x√1+y+y√1+x=0, then (dy/dx)=

Question

If x√1+y+y√1+x=0, then (dy/dx)= (A) (x+1/x) (B) (1/1+x) (C) (-1/(1+x)2) (D) (x/1+x)

Tardigrade Admin · Accepted Answer

x √1+y+y √1+x=0 =x √1+y=-y √1+x Squaring, x2(1+y)=y2(1+x) =x2+x2 y=y2+x y2 =x2-y2=x y2-x2 y =x2-y2=x y(y-x) =(x +y)(x-y)=x y(y-x) =(x +y)=-x y =x+ x y+ y=0 =(1+x) y=-x y=-x /(1+x) differentiating, (d y/d x)=((1+x)-x/(1+x)2) =(-1/(1+x)2)

Q. If $x 1 + y + y 1 + x = 0,$ then $\frac{d y}{d x} =$

$\frac{x + 1}{x}$

$\frac{1}{1 + x}$

$\frac{- 1}{( 1 + x ) ^{2}}$

$\frac{x}{1 + x}$

Q. If x1+y​+y1+x​=0, then dxdy​=

xx+1​

1+x1​

(1+x)2−1​

1+xx​

x1+y​+y1+x​=0 =x1+y​=−y1+x​ Squaring, x2(1+y)=y2(1+x) =x2+x2y=y2+xy2 =x2−y2=xy2−x2y =x2−y2=xy(y−x) =(x+y)(x−y)=xy(y−x) =(x+y)=−xy =x+xy+y=0 =(1+x)y=−x y=−x/(1+x) differentiating, dxdy​=(1+x)2(1+x)−x​ =(1+x)2−1​

Q. If $x 1 + y + y 1 + x = 0,$ then $\frac{d y}{d x} =$

$\frac{x + 1}{x}$

$\frac{1}{1 + x}$

$\frac{- 1}{( 1 + x ) ^{2}}$

$\frac{x}{1 + x}$