Question

If $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$, then$\dfrac{{dy}}{{dx}}$ is equal to
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 shall analyze the given information so that we are able to solve the problem. Generally in Mathematics, the derivative refers to the rate of change of a function with respect to a variable. First, we need to solve the given equation to obtain$y$. Here, we are applying the quotient rule to find the required answer
 Here we are asked to find the derivative of$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$.
Formula to be used:
 The formula that is applied in the quotient rule of differentiation is as follows.
                                   $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}\left( u \right) - u\dfrac{d}{{dx}}\left( v \right)}}{{{{\left( v \right)}^2}}}$

Complete step by step answer:
It is given that$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$
$ \Rightarrow x\sqrt {1 + y} = - y\sqrt {1 + x} $
We shall take squares on both sides.
\[ \Rightarrow {\left( {x\sqrt {1 + y} } \right)^2} = {\left( { - y\sqrt {1 + x} } \right)^2}\]
\[ \Rightarrow {x^2}{\left( {\sqrt {1 + y} } \right)^2} = {y^2}{\left( {\sqrt {1 + x} } \right)^2}\]
\[ \Rightarrow {x^2}\left( {1 + y} \right) = {y^2}\left( {1 + x} \right)\]
\[ \Rightarrow {x^2} + {x^2}y = {y^2} + x{y^2}\]
\[ \Rightarrow {x^2} - {y^2} = x{y^2} - {x^2}y\]
\[ \Rightarrow {x^2} - {y^2} = xy\left( {y - x} \right)\]
\[ \Rightarrow \left( {x - y} \right)\left( {x + y} \right) = xy\left( {y - x} \right)\] (Here we applied the formula${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ )
\[ \Rightarrow \left( {x - y} \right)\left( {x + y} \right) = - xy\left( {x - y} \right)\]
\[ \Rightarrow \left( {x + y} \right) = - xy\] (Here the term$x - y$gets canceled on both sides)
\[ \Rightarrow xy + y = - x\]
\[ \Rightarrow \left( {x + 1} \right)y = - x\]
\[ \Rightarrow y = - \dfrac{x}{{1 + x}}\]
Here in this question, we are asked to calculate $\dfrac{{dy}}{{dx}}$
Hence, we shall differentiate \[y = - \dfrac{x}{{1 + x}}\] with respect to$x$ .
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{x}{{1 + x}}} \right)$
      $ =- \dfrac{{\left( {1 + x} \right)\dfrac{d}{{dx}}\left( x \right) - x\dfrac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}$
(Here we applied the formula of the quotient rule of differentiation$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}\left( u \right) - u\dfrac{d}{{dx}}\left( v \right)}}{{{{\left( v \right)}^2}}}$ )
$ \Rightarrow \dfrac{{dy}}{{dx}} = -\dfrac{{\left( {1 + x} \right)1 - x \times 1}}{{{{\left( {1 + x} \right)}^2}}}$
            $ = -\dfrac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}$
            $ = \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}$
Hence, $\dfrac{{dy}}{{dx}}$$ = \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}$

So, the correct answer is “Option B”.

Note: If we are asked to calculate the derivative of a given equation, we need to first analyze the given problem where we are able to apply the derivative formulae and the derivative refers to the rate of change of a function with respect to a variable.
          Here, we have applied the quotient rule of derivative formulae that are needed to obtain the desired answer. Hence, we got$\dfrac{{dy}}{{dx}}$$ = \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}$.
      Generally, we use the quotient rule when the variables are in the form of a ratio. That is it is a formula applied in the differentiation when one function is divided by another function.