If x √(1+y)+y √(1+x)=0, then (d y/d x) equals -

Question

If x √(1+y)+y √(1+x)=0, then (d y/d x) equals - (A) (1/(1+x)2) (B) -(1/(1+x)2) (C) -(1/(1+x))+(1/(1+x)2) (D) none of these

Tardigrade Admin · Accepted Answer

x2(1+y)=y2(1+x) ⇒ x2-y2+x y(x-y)=0 ⇒ (x-y)(x+y+x y)=0 Now x ≠ y [does not satisfy the given equation] ∴ x+y+x y=0 ⇒ y=(-x/1+x) ∴ (d y/d x)=(-(1+x)+x/(1+x)2)=-(1/(1+x)2)

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

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

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

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

none of these

Q. If x(1+y)​+y(1+x)​=0, then dxdy​ equals -

(1+x)21​

−(1+x)21​

−(1+x)1​+(1+x)21​

none of these

x2(1+y)=y2(1+x) ⇒x2−y2+xy(x−y)=0 ⇒(x−y)(x+y+xy)=0 Now x=y [does not satisfy the given equation] ∴x+y+xy=0⇒y=1+x−x​ ∴dxdy​=(1+x)2−(1+x)+x​=−(1+x)21​

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

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

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

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