Question

1.If \[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\] , then prove that \[\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}(x \ne y)\].
2. If \[\cos y = x.\cos (a + y)\],then prove that \[\dfrac{{dy}}{{dx}} = \dfrac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}\].

Answer 1

Hint:
It’s obvious that we have to perform differentiation in this solution. First we’ll try to seperate the independent variable that is x and the dependent variable that is y. Then we’ll perform differentiation. If we can’t separate the variables then we’ll use the product rule of differentiation.

Complete step by step solution:
1. \[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\]

Formula used:
1. \[{a^2} - {b^2} = (a + b)(a - b)\]
2. \[\dfrac{u}{v} = \dfrac{{u'v - v'u}}{{{v^2}}}\]
Given that,
\[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\]
\[ \Rightarrow x\sqrt {1 + y} = - y\sqrt {1 + x} \]
Squaring both sides,
\[
   \Rightarrow {x^2}\left( {1 + y} \right) = {\left( { - y} \right)^2}(1 + x) \\
   \Rightarrow {x^2}\left( {1 + y} \right) = {y^2}(1 + x) \\
   \Rightarrow {x^2} + {x^2}y = {y^2} + {y^2}x \\
   \Rightarrow {x^2} - {y^2} = {y^2}x - {x^2}y \\
 \]
Using formula , \[{a^2} - {b^2} = (a + b)(a - b)\]
\[ \Rightarrow (x + y)(x - y) = - {x^2}y + {y^2}x\]
\[ \Rightarrow (x + y)(x - y) = - xy(x - y)\] ..taking –xy common on right side.
\[
   \Rightarrow x + y = - xy \\
   \Rightarrow x = - xy - y \\
   \Rightarrow x = - y(x + 1) \\
   \Rightarrow y = \dfrac{{ - x}}{{x + 1}} \\
 \]
Using quotient’s rule, \[\dfrac{u}{v} = \dfrac{{u'v - v'u}}{{{v^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{d}{{dx}}( - x) \times (x + 1) - \dfrac{d}{{dx}}(x + 1) \times ( - x)}}{{{{(x + 1)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{( - 1)(x + 1) - (1)( - x)}}{{{{(x + 1)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - x - 1 + x}}{{{{(x + 1)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{{(x + 1)}^2}}}\]
Hence proved.

2. \[\cos y = x.\cos (a + y)\]

Formula used:
1. \[\dfrac{u}{v} = \dfrac{{u'v - v'u}}{{{v^2}}}\]
2. \[\sin (x - y) = \sin x.\cos y - \cos x.\sin y\]

Given that,
\[\cos y = x.\cos (a + y)\]
\[
   \Rightarrow x = \dfrac{{\cos y}}{{\cos (a + y)}} \\
   \Rightarrow \dfrac{d}{{dx}}x = \dfrac{d}{{dx}}\dfrac{{\cos y}}{{\cos (a + y)}} \\
   \Rightarrow 1 = \dfrac{d}{{dx}}\dfrac{{\cos y}}{{\cos (a + y)}}\dfrac{{dy}}{{dy}} \\
   \Rightarrow 1 = \dfrac{d}{{dy}}\dfrac{{\cos y}}{{\cos (a + y)}}\dfrac{{dy}}{{dx}} \\
 \]
Using quotient’s rule, \[\dfrac{u}{v} = \dfrac{{u'v - v'u}}{{{v^2}}}\]
\[
   \Rightarrow 1 = \left[ {\dfrac{{\cos (a + y)\dfrac{d}{{dy}}\cos y - (\cos y\dfrac{d}{{dy}}\cos (a + y))}}{{{{\left( {\cos \left( {a + y} \right)} \right)}^2}}}} \right]\dfrac{{dy}}{{dx}} \\
   \Rightarrow 1 = \left[ {\dfrac{{\cos (a + y)( - \sin y) - (\cos y( - \sin (a + y))\dfrac{d}{{dy}}(a + y))}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}} \\
   \Rightarrow 1 = \left[ {\dfrac{{\cos (a + y)( - \sin y) - (\cos y( - \sin (a + y)))1}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}} \\
   \Rightarrow 1 = \left[ {\dfrac{{\cos (a + y)( - \sin y) + \cos y\sin (a + y)}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}} \\
 \]
Rearranging the numerator terms
\[ \Rightarrow 1 = \left[ {\dfrac{{\sin (a + y)\cos y - \cos (a + y)\sin y}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}}\]
Using formula \[\sin (x - y) = \sin x.\cos y - \cos x.\sin y\]
\[
   \Rightarrow 1 = \left[ {\dfrac{{\sin ((a + y) - y)}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}} \\
   \Rightarrow 1 = \left[ {\dfrac{{\sin a}}{{{{\cos }^2}(a + y)}}} \right]\dfrac{{dy}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\cos }^2}(a + y)}}{{\sin a}} \\
 \]
Hence proved.

Note:
Here the expectation from us was to differentiate. The same question can be asked from differential equation chapter. they’ll as constants one or two and ask to form differential equation. In such situation we need to focus on removing the constant from the equation by differentiating it.