Question

If
$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$
Prove that
$\frac{dy}{dx} = \frac{ - 1}{( {1 + x)}^{2} }$
​

Answer 1

First find Y , by solving the given equation

we get

Y= -x/1+x

And at last ,

Differentiate Y wrt to X

Formula used :

$\red{\huge \bold{\frac{d}{dx} ( \frac{u}{v} ) = \frac{v \: \frac{d}{dx} u - u \frac{d}{dx} v}{ {v}^{2} } }}$

Solution refer to the attachment

Answer 2

Given :---

• x√(1+y) + y√(1+x) = 0

To Prove :--

• dy/dx = -1/(1+x)²

Solution :---

→ x√(1+y) + y√(1+x) = 0

→ x√(1+y) = - y√(1+x)

Squaring both sides we get,

→ x² * (1+y) = y² (1+x)

→ x² + x²y = y² + y²x

→ x² - y² = y²x - x²y

using (a²-b²) = (a+b)(a-b) in LHS and, taking xy common From RHS , we get,

→ (x-y)(x+y) = xy(y-x)

Taking (-1) common From RHS,

→ (x-y)(x+y) = - xy(x-y)

(x-y) will cancel from both sides now,

→ (x+y) = (-xy)

→ x = - xy - y

Taking (-y) common From RHS now,

→ x = -y(x+1)

Cross - Multiply now,

→ y = -(x/(x+1))

____________________________

Now, differentiating both sides with respect to x, by using Product rule of Differentiate , we get,

→ d/dx [f(x) * g(x) ] = f(x) * d/dx[g(x)] + g(x) * d/dx[f(x)]

→ dy = - [ (x+1)*d/dx[x] - x * d/dx[x+1] ]

dx (x+1)²

→ dy = - [ (x+1)*1 - x * 1 ]

dx (x+1)²

→ dy = - [ x+1 - x ]

dx (x+1)²

→ dy = -1

dx (x+1)²

✪✪ Hence Proved ✪✪