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Question: If \(x\sqrt {1 + y} + y\sqrt {1 + x} = 0\), then\(\dfrac{{dy}}{{dx}}\) is equal to A. \(\dfrac{1}{...

If x1+y+y1+x=0x\sqrt {1 + y} + y\sqrt {1 + x} = 0, thendydx\dfrac{{dy}}{{dx}} is equal to
A. 1(1+x)2\dfrac{1}{{{{\left( {1 + x} \right)}^2}}}
B. −1(1+x)2 - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}
C. 1(1+x2)\dfrac{1}{{\left( {1 + {x^2}} \right)}}
D. 1(1−x2)\dfrac{1}{{\left( {1 - {x^2}} \right)}}

Explanation

Solution

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 obtainyy. Here, we are applying the quotient rule to find the required answer
Here we are asked to find the derivative ofx1+y+y1+x=0x\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.
ddx(uv)=vddx(u)−uddx(v)(v)2\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 thatx1+y+y1+x=0x\sqrt {1 + y} + y\sqrt {1 + x} = 0
⇒x1+y=−y1+x\Rightarrow x\sqrt {1 + y} = - y\sqrt {1 + x}
We shall take squares on both sides.
⇒(x1+y)2=(−y1+x)2\Rightarrow {\left( {x\sqrt {1 + y} } \right)^2} = {\left( { - y\sqrt {1 + x} } \right)^2}
⇒x2(1+y)2=y2(1+x)2\Rightarrow {x^2}{\left( {\sqrt {1 + y} } \right)^2} = {y^2}{\left( {\sqrt {1 + x} } \right)^2}
⇒x2(1+y)=y2(1+x)\Rightarrow {x^2}\left( {1 + y} \right) = {y^2}\left( {1 + x} \right)
⇒x2+x2y=y2+xy2\Rightarrow {x^2} + {x^2}y = {y^2} + x{y^2}
⇒x2−y2=xy2−x2y\Rightarrow {x^2} - {y^2} = x{y^2} - {x^2}y
⇒x2−y2=xy(y−x)\Rightarrow {x^2} - {y^2} = xy\left( {y - x} \right)
⇒(x−y)(x+y)=xy(y−x)\Rightarrow \left( {x - y} \right)\left( {x + y} \right) = xy\left( {y - x} \right) (Here we applied the formulaa2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) )
⇒(x−y)(x+y)=−xy(x−y)\Rightarrow \left( {x - y} \right)\left( {x + y} \right) = - xy\left( {x - y} \right)
⇒(x+y)=−xy\Rightarrow \left( {x + y} \right) = - xy (Here the termx−yx - ygets canceled on both sides)
⇒xy+y=−x\Rightarrow xy + y = - x
⇒(x+1)y=−x\Rightarrow \left( {x + 1} \right)y = - x
⇒y=−x1+x\Rightarrow y = - \dfrac{x}{{1 + x}}
Here in this question, we are asked to calculate dydx\dfrac{{dy}}{{dx}}
Hence, we shall differentiate y=−x1+xy = - \dfrac{x}{{1 + x}} with respect toxx .
dydx=ddx(−x1+x)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{x}{{1 + x}}} \right)
=−(1+x)ddx(x)−xddx(1+x)(1+x)2=- \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 differentiationddx(uv)=vddx(u)−uddx(v)(v)2\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}}} )
⇒dydx=−(1+x)1−x×1(1+x)2\Rightarrow \dfrac{{dy}}{{dx}} = -\dfrac{{\left( {1 + x} \right)1 - x \times 1}}{{{{\left( {1 + x} \right)}^2}}}
=−1+x−x(1+x)2= -\dfrac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}
=−1(1+x)2= \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}
Hence, dydx\dfrac{{dy}}{{dx}} =−1(1+x)2 = \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 gotdydx\dfrac{{dy}}{{dx}} =−1(1+x)2 = \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.