Question

If x√1+y+y√1+x=0,then dy\dx is equal to​

Answer 1

Answer:

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

Step-by-step explanation:

Given that,

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

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

On squaring both sides, we get

$\sf \: (x \sqrt{1 + y})^{2} = (- y \sqrt{1 + x})^{2}$

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

$\sf \: {x}^{2} + y {x}^{2} = {y}^{2} + x {y}^{2}$

$\sf \: {x}^{2} - {y}^{2} = x {y}^{2} - {x}^{2}y$

$\sf \: (x + y)(x - y)= xy(y - x)$

$\sf \: (x + y)(x - y)= - xy(x - y)$

$\sf \: x + y= - xy$

$\sf \: x= - xy - y$

$\sf \: x= - y(x + 1)$

$\sf \: y = - \: \dfrac{x}{x + 1}$

On differentiating both sides w. r. t. x, we get

$\sf \: \dfrac{d}{dx} y =\dfrac{d}{dx}\left( \dfrac{ - x}{x + 1}\right)$

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

$\sf \: \dfrac{dy}{dx} = \dfrac{(x + 1)( - 1) + x(1 + 0)}{ {(x + 1)}^{2} }$

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

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

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Formulae used:

$\sf \: \dfrac{d}{dx}\dfrac{u}{v} = \dfrac{u\dfrac{d}{dx}v - v\dfrac{d}{dx}u}{ {v}^{2} }$

$\sf \: \dfrac{d}{dx}x = 1$

$\sf \: \dfrac{d}{dx}k = 0$

Answer 2

Answer:

Given that,

\begin{gathered}\sf \: x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \\ \\ \end{gathered}

x

1+y

+y

1+x

=0

\begin{gathered}\sf \: x \sqrt{1 + y} = - y \sqrt{1 + x} \\ \\ \end{gathered}

x

1+y

=−y

1+x

On squaring both sides, we get

\begin{gathered}\sf \: (x \sqrt{1 + y})^{2} = (- y \sqrt{1 + x})^{2} \\ \\ \end{gathered}

(x

1+y

)

2

=(−y

1+x

)

2

\begin{gathered}\sf \: {x}^{2}(1 + y) = {y}^{2}(1 + x) \\ \\ \end{gathered}

x

2

(1+y)=y

2

(1+x)

\begin{gathered}\sf \: {x}^{2} + y {x}^{2} = {y}^{2} + x {y}^{2} \\ \\ \end{gathered}

x

2

+yx

2

=y

2

+xy

2

\begin{gathered}\sf \: {x}^{2} - {y}^{2} = x {y}^{2} - {x}^{2}y \\ \\ \end{gathered}

x

2

−y

2

=xy

2

−x

2

y

\begin{gathered}\sf \: (x + y)(x - y)= xy(y - x) \\ \\ \end{gathered}

(x+y)(x−y)=xy(y−x)

\begin{gathered}\sf \: (x + y)(x - y)= - xy(x - y) \\ \\ \end{gathered}

(x+y)(x−y)=−xy(x−y)

\begin{gathered}\sf \: x + y= - xy \\ \\ \end{gathered}

x+y=−xy

\begin{gathered}\sf \: x= - xy - y \\ \\ \end{gathered}

x=−xy−y

\begin{gathered}\sf \: x= - y(x + 1) \\ \\ \end{gathered}

x=−y(x+1)

\begin{gathered}\sf \: y = - \: \dfrac{x}{x + 1} \\ \\ \end{gathered}

y=−

x+1

x

On differentiating both sides w. r. t. x, we get

\begin{gathered}\sf \: \dfrac{d}{dx} y =\dfrac{d}{dx}\left( \dfrac{ - x}{x + 1}\right) \\ \\ \end{gathered}

dx

d

y=

dx

d

(

x+1

−x

)

\begin{gathered}\sf \: \dfrac{dy}{dx} = \dfrac{(x + 1)\dfrac{d}{dx}( - x) - ( - x)\dfrac{d}{dx}(x + 1)}{ {(x + 1)}^{2} } \\ \\ \end{gathered}

dx

dy

=

(x+1)

2

(x+1)

dx

d

(−x)−(−x)

dx

d

(x+1)

\begin{gathered}\sf \: \dfrac{dy}{dx} = \dfrac{(x + 1)( - 1) + x(1 + 0)}{ {(x + 1)}^{2} } \\ \\ \end{gathered}

dx

dy

=

(x+1)

2

(x+1)(−1)+x(1+0)

\begin{gathered}\sf \: \dfrac{dy}{dx} = \dfrac{ - x - 1 + x}{ {(x + 1)}^{2} } \\ \\ \end{gathered}

dx

dy

=

(x+1)

2

−x−1+x

\begin{gathered}\implies\sf \: \sf \: \dfrac{dy}{dx} = \dfrac{ - 1}{ {(x + 1)}^{2} } \\ \\ \end{gathered}

⟹

dx

dy

=

(x+1)

2

−1

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