Question

answer the question I will mark u as brain list​

Answer 1

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER :

• $When \: u = f(x)\: and\:v = g(x) then \\\dfrac{d}{dx}\, \dfrac{u}{v}= \dfrac{vu'-uv'}{v^2}$

where,
$u' = \dfrac{du}{dx} \: and\: v' = \dfrac{dv}{dx}$

•$(a^2-b^2) = (a+b)(a-b)$

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¶¶¶ SOLUTION:

Given,
$\implies x\sqrt{1+y}+y\sqrt{1+x}=0$

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

$\implies x^2 \cdot (1+y) = (-y)^2 \cdot (1+x)$

$\implies x^2 \cdot (1+y) = y^2 \cdot (1+x)$

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

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

$\implies (x+y)(x-y) = -xy \cdot (x-y)$

$\implies x+y = -xy$

$\implies x = -xy-y$

$\implies x = -y \cdot (x+1)$

$\implies y = \dfrac{-x}{x+1}$

Differentiate w.r.t x on both sides

Apply u/v rule on RHS

$\implies \dfrac{dy}{dx} = -\dfrac{(x+1) \cdot -1 - (-x) \cdot 1}{(x+1)^2}$

$\implies \dfrac{dy}{dx} = \dfrac{-1}{(x+1)^2}$

This answer is in Option -2
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Answer 2

Step-by-step explanation:

Nice Explanation Vemugantu Rahul