Question

if x(1+x)^1/2+y(1+x)^1/2=0 then (1+x)^2dy/dx
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Answer 1

Step-by-step explanation:

given

x√(1+y)+y√(1+x)=0

x√(1+y) = - y√(1+x)

squareing on both sides

x²(1+y) = - y²(1+x)

x²+x²y = -y²- y²x

x²-y² = y²x-x²y

(x+y)(x-y)=-xy(x-y)

x+y=-xy

case1

x= -y(x+1)---------i

case2

y=-x/(X+1)---------i

according to problem

$\frac{dy}{dx} = \frac{ - 1(x + 1) - 1( - x)}{ {(x + 1)}^{2} } \\ \\ on \: solving \\ \frac{dy}{dx} = \frac { - 1}{ {( x + 1)}^{2} } \\ \\ (x + 1) ^{2} \frac{dy}{dx} = - 1$

on solving