If x -1-y-y -1-x-0 then the value of x-12d2 y-d x2-2x-1 d y-d x at y-3 is equal to

X√(1+y)= y√(1+x) x^2(1+y)=y^2(1+x) x^2+x^2y=y^2+y^2x x^2 y^2=y^2x x^2y (x+y)(x y)=xy(y x) x+y= xy x= y(x+1) y= x/x+1 dy/dx= (x+1) x/(x+1)^2= 1/(x+1)^2 (x+1)^2dy ...

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Byju's Answer

Standard XII

Mathematics

Properties of Conjugate of a Complex Number

If x √1+y+y √...

Question

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ then the value of $(x + 1)^{2} \frac{d^{2} y}{d x^{2}} + 2 (x + 1) \frac{d y}{d x}$ at $y = 3$ is equal to

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Solution

$x \sqrt{1 + y} = - y \sqrt{1 + x}$
$x^{2} (1 + y) = y^{2} (1 + x)$
$x^{2} + x^{2} y = y^{2} + y^{2} x$
$x^{2} - y^{2} = y^{2} x - x^{2} y$
$(x + y) (x - y) = x y (y - x)$
$x + y = - x y$
$x = - y (x + 1)$
$y = - \frac{x}{x + 1}$
$\frac{d y}{d x} = - \frac{(x + 1) - x}{(x + 1)^{2}} = - \frac{1}{(x + 1)^{2}}$
$(x + 1)^{2} \frac{d y}{d x} = - 1$
$(x + 1)^{2} \frac{d^{2} y}{d x^{2}} + 2 (x + 1) \frac{d y}{d x} = 0$

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If x√1+y+y√1+x=0 then the value of (x+1)2d2ydx2+2(x+1)dydx at y=3 is equal to

x√1+y=−y√1+x x2(1+y)=y2(1+x) x2+x2y=y2+y2x x2−y2=y2x−x2y (x+y)(x−y)=xy(y−x) x+y=−xy x=−y(x+1) y=−xx+1 dydx=−(x+1)−x(x+1)2=−1(x+1)2 (x+1)2dydx=−1 (x+1)2d2ydx2+2(x+1)dydx=0

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ then the value of $(x + 1)^{2} \frac{d^{2} y}{d x^{2}} + 2 (x + 1) \frac{d y}{d x}$ at $y = 3$ is equal to