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# Solution for If X √ 1 + Y + Y √ 1 + X = 0 , Prove that ( 1 + X ) 2 D Y D X + 1 = 0 ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ , prove that $\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0$  ?

#### Solution

$\text { We have }, x\sqrt{1 + y} + y\sqrt{1 + x} = 0$

$\Rightarrow x\sqrt{1 + y} = - y\sqrt{1 + x}$

$\text{ Squaring both sides, we get } ,$

$\Rightarrow \left( x\sqrt{1 + y} \right)^2 = \left( - y\sqrt{1 + x} \right)^2$

$\Rightarrow x^2 \left( 1 + y \right) = y^2 \left( 1 + x \right)$

$\Rightarrow x^2 + x^2 y = y^2 + y^2 x$

$\Rightarrow x^2 - y^2 = y^2 x - x^2 y$

$\Rightarrow \left( x - y \right)\left( x + y \right) = xy\left( y - x \right)$

$\Rightarrow \left( x + y \right) = - xy$

$\Rightarrow y + xy = - x$

$\Rightarrow y\left( 1 + x \right) = - x$

$\Rightarrow y = \frac{- x}{\left( 1 + x \right)}$

Differentiating with respect to x, we get,

$\Rightarrow \frac{d y}{d x} = \left[ \frac{- \left( 1 + x \right)\frac{d}{dx}\left( x \right) - \left( - x \right)\frac{d}{dx}\left( x + 1 \right)}{\left( 1 + x \right)^2} \right]$

$\Rightarrow \frac{d y}{d x} = \left[ \frac{- \left( 1 + x \right)\left( 1 \right) + x\left( 1 \right)}{\left( 1 + x \right)^2} \right]$

$\Rightarrow \frac{d y}{d x} = \left[ \frac{- 1 - x + x}{\left( 1 + x \right)^2} \right]$

$\Rightarrow \frac{d y}{d x} = \frac{- 1}{\left( 1 + x \right)^2}$

$\Rightarrow \left( 1 + x \right)^2 \frac{d y}{d x} = - 1$

$\Rightarrow \left( 1 + x \right)^2 \frac{d y}{d x} + 1 = 0$

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Solution If X √ 1 + Y + Y √ 1 + X = 0 , Prove that ( 1 + X ) 2 D Y D X + 1 = 0 ? Concept: Simple Problems on Applications of Derivatives.
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