Question

if x√(1+y) + y√(1+x)=0 find dy/dx

Answer 1

$\textbf{Hello Mate}$

$\textbf{Your Answer}$>>>

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

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

$\textbf{Now Squaring Both Sides}$>>>

$\upper{x}^{2}$ ( 1 + y ) = $\upper{y}^{2}$ ( 1 + x )

$\upper{x}^{2}$ ( 1 + y ) - $\upper{y}^{2}$ ( 1 + x ) = 0

$\upper{x}^{2}$ + $\upper{x}^{2}$y - $\upper{y}^{2}$ - $\upper{y}^{2}$x = 0

$\textbf{Now Arranging}$>>>

$\upper{x}^{2}$ - $\upper{y}^{2}$ + $\upper{x}^{2}$y - $\upper{y}^{2}$x = 0

$\textbf{We Know that}$

$\upper{a}^{2}$ - $\upper{b}^{2}$ = ( a - b )( a + b )

$\textbf{Applying Formulla}$>>>

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

$\textbf{ Taking x- y common}$>>>

( x - y ) [ x + y + xy ] = 0

x + y + xy = 0

x + y ( 1 + x ) = 0

y ( 1 + x ) = - x

y = $\frac{- x}{1 + x}$

$\textbf{Now Differentiate Both Sides wrt. x}$

$\textbf{So we get }$>>>

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

Be Brainly..

@karangrover12