If x(1+y)^1/2+y(1+x)^1/2=0 then dy/dx is ?
Answers
Answered by
5
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 ]
Answered by
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 ❤
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