If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ $xsqrt(1 + y) + ysqrt(1 + x) = 0$ . Then how will you prove that $\frac{d y}{d x} = - \frac{1}{{(1 + x)}^{2}}$ $dy/dx = -1/(1 + x)^2$ ??

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Answer 1 · 2017-07-24T17:23:31

$\frac{d y}{d x} = - \frac{1}{{(x + 1)}^{2}}$ $dy/dx=-1/(x+1)^2$

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ $xsqrt(1 + y) + ysqrt(1 + x) = 0$

$x \sqrt{1 + y} = - y \sqrt{1 + x}$ $xsqrt(1 + y)=-ysqrt(1 + x)$

$x^{2} \cdot (1 + y) = {(- y)}^{2} \cdot (1 + x)$ $x^2*(1 + y)=(-y)^2*(1 + x)$

$x^{2} \cdot (1 + y) = y^{2} \cdot (1 + x)$ $x^2*(1 + y)=y^2*(1 + x)$

$x^{2} + x^{2} \cdot y = y^{2} + y^{2} \cdot x$ $x^2 + x^2*y=y^2+y^2*x$

$x^{2} - y^{2} = y^{2} \cdot x - x^{2} \cdot y$ $x^2 -y^2=y^2*x-x^2*y$

$(x + y) \cdot (x - y) = - x y \cdot (x - y)$ $(x+y) * (x-y) = -xy * (x-y)$

$x + y = - x y$ $x+y = -xy$

$x = - x y - y$ $x = -xy-y$

$x = - y \cdot (x + 1)$ $x = -y* (x+1)$

$y = - \frac{x}{x + 1}$ $y=-x/(x+1)$

$\frac{d y}{d x} = \frac{- 1 \cdot (x + 1) - (- x) \cdot 1}{{(x + 1)}^{2}}$ $dy/dx=[-1*(x+1)-(-x)*1]/(x+1)^2$

$\frac{d y}{d x} = - \frac{1}{{(x + 1)}^{2}}$ $dy/dx=-1/(x+1)^2$

1) I solved this equation for y.

2) I differentiated both sides.

If xsqrt(1 + y) + ysqrt(1 + x) = 0x√1+y+y√1+x=0xsqrt(1 + y) + ysqrt(1 + x) = 0. Then how will you prove that dy/dx = -1/(1 + x)^2dydx=−1(1+x)2dy/dx = -1/(1 + x)^2??