Question

If x(1+y)^1/2+y(1+x)^1/2=0 then dy/dx is ?​

Answer 1

Answer:

Step-by-step explanation:

x (1+y)^(1/2) = -y (1+x)^(1/2)  ------ (a)

SQUARING

(x^2)(1+y) = (y^2)(1+x)

x^2 + y(x^2) - y^2 - x(y^2) = 0

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(x-y)(x+y) + xy(x-y) = 0

(x-y)(x+y+xy) = 0

x=y doesn't satisfy (a)

So

x+y+xy = 0

y = -x / (1+x)

Differentiating w.r.t x

(dy/dx) = -1 / [ (1+x)^2 ]

Answer 2

Answer:

squaring the given equation

(x^2)(1+y)=(y^2)(1+x)

(x^2)(yx^2)-(y^2)-x(y^2)=0

(x-y)(x+y) +xy(x-y)=0

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(x-y)(x+y+xy)=0

x=y(Not satisfy given eq.)

so x+y +xy =0

y=-x/(1+x)

So,

dy/dX= -1/[(1+x^2)]

Hope helps ❤