If x{1+y}+y{1+x}=0, then \;dy/dx is equal to

Correct Answer is : − 1(1+x)2-\;1/(1+x)²−(1+x)21

Correct Answer is : − 1(1+x)2-\;1/(1+x)²−(1+x)21

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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

Question : 146 of 160

Marks: +1, -0

x\sqrt{1+y}+y\sqrt{1+x}=0

\;\frac{dy}{dx}

Solution:

Given that

\;\;x\sqrt{1+y}+y\sqrt{1+x}=0

. . . (i)

\;\text{ or }\;\;\;\;x\sqrt{1+y}=-y\sqrt{1+x}

On squaring both sides, we get

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

\Rightarrow x^2-y^2+x^2y-xy^2=0

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

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

x-y \neq 0

because it does not satisfy the Eq. (i)

\;\therefore \;\;\;x+y+xy=0

\;\Rightarrow \;\;\;y=-\;\frac{x}{1+x}

\;

On differentiating w.r.t.

x

, we get

\;\frac{dy}{dx}\;=-\;\frac{(1+x)(1)-x(1)}{(1+x)^2}

\;\;=-\;\frac{1}{(1+x)^2}

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