Question

If \[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\], then find \[\dfrac{{dy}}{{dx}}\].
A. \[\dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]
B. \[ - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]
C. \[\dfrac{1}{{\left( {1 + {x^2}} \right)}}\]
D. \[\dfrac{1}{{\left( {1 - {x^2}} \right)}}\]

Answer 1

Hint:First we will rewrite the equation as \[x\sqrt {1 + y} = - y\sqrt {1 + x} \]. Then taking square both sides of the equation to remove the square root. Then simplify the equation and find \[y\] in terms of \[x\]. Then we will apply the quotient formula \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\] to get \[\dfrac{{dy}}{{dx}}\].

Formula Used:
\[{\left( {ab} \right)^m} = {a^m}{b^m}\]
\[{\left( {\sqrt x } \right)^2} = x\]
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Quotient formula: \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]

Complete step by step solution:
Given equation is
\[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\]
Rewrite the above equation:
\[x\sqrt {1 + y} = - y\sqrt {1 + x} \]
Taking square root both sides of the equation
\[ \Rightarrow {\left( {x\sqrt {1 + y} } \right)^2} = {\left( { - y\sqrt {1 + x} } \right)^2}\]
Apply the formula \[{\left( {ab} \right)^m} = {a^m}{b^m}\]
\[ \Rightarrow {x^2}{\left( {\sqrt {1 + y} } \right)^2} = {y^2}{\left( {\sqrt {1 + x} } \right)^2}\]
Apply the formula \[{\left( {\sqrt x } \right)^2} = x\]
\[ \Rightarrow {x^2}\left( {1 + y} \right) = {y^2}\left( {1 + x} \right)\]
Apply distributive property
\[ \Rightarrow {x^2} + {x^2}y = {y^2} + {y^2}x\]
\[ \Rightarrow {x^2} - {y^2} = {y^2}x - {x^2}y\]
Apply the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] on the left side
\[ \Rightarrow \left( {x - y} \right)\left( {x + y} \right) = {y^2}x - {x^2}y\]
Take common \[xy\] from the right side
\[ \Rightarrow \left( {x - y} \right)\left( {x + y} \right) = xy\left( {y - x} \right)\]
\[ \Rightarrow \left( {x - y} \right)\left( {x + y} \right) = - xy\left( {x - y} \right)\]
Cancel out \[\left( {x - y} \right)\]
\[ \Rightarrow \left( {x + y} \right) = - xy\]
\[ \Rightarrow x = - xy - y\]
\[ \Rightarrow x = - y\left( {1 + x} \right)\]
\[ \Rightarrow y = - \dfrac{x}{{1 + x}}\]
Differentiate with respect to \[x\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {1 + x} \right)\dfrac{d}{{dx}}x - x\dfrac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {1 + x} \right) \cdot 1 - x \cdot 1}}{{{{\left( {1 + x} \right)}^2}}}\]
Apply the formula \[\dfrac{{dx}}{{dx}} = 1\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Hence option B is the correct option.

Note:Sometimes students apply the derivative formula in the give equation \[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\]. That is a length process and they are unable to reach the desired result. So first we have to simplify the given equation then find the derivative of it.