Question

If $y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } $ , then show that $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{2{y^3} - 2xy - 1}}$.

Answer 1

Hint: In order to solve the equation, check that where the term $y$ has repeated and, then mark them as $y$ , we will get a shorter equation. Then to remove the square root, square both the sides. Then differentiate the terms with respect to $x$ and separate the variables on one side.

Formula used:
1. \[\dfrac{{d{y^n}}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}\]
2. \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
3. \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + v\dfrac{{d\left( u \right)}}{{dx}}\]

Complete step by step solution:
We are given the equation $y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } $.
We can see that after the second root, from the third root $y$ is getting repeated, as $y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } $.
So, writing $\sqrt {x + \sqrt {y + .........\infty } } $ as $y$ , we get the equation as:
$
  y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } \\
   \Rightarrow y = \sqrt {x + \sqrt {y + y} } \\
 $
It can be written as:
$
  y = \sqrt {x + \sqrt {y + y} } \\
   \Rightarrow y = \sqrt {x + \sqrt {2y} } \\
 $
Squaring both the sides, in order to remove the square root, as we know that ${\left( {\sqrt x } \right)^2} = x$.
So,
 $
  y = \sqrt {x + \sqrt {2y} } \\
   \Rightarrow {\left( y \right)^2} = {\left( {\sqrt {x + \sqrt {2y} } } \right)^2} \\
   \Rightarrow {y^2} = x + \sqrt {2y} \\
 $
Subtracting both the sides of the above equation by $x$:
$
  {y^2} = x + \sqrt {2y} \\
   \Rightarrow {y^2} - x = x + \sqrt {2y} - x \\
   \Rightarrow {y^2} - x = \sqrt {2y} \\
 $

Since, we have one more square root on the right side, so squaring both the sides, and we get:
$
  {y^2} - x = \sqrt {2y} \\
   \Rightarrow {\left( {{y^2} - x} \right)^2} = {\left( {\sqrt {2y} } \right)^2} \\
   \Rightarrow {\left( {{y^2} - x} \right)^2} = 2y \\
 $
Opening the brackets using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
$
  {\left( {{y^2} - x} \right)^2} = 2y \\
   = > {y^4} + {x^2} - 2x{y^2} = 2y \\
 $
Subtracting both the sides by $2y$ , we get:
$
  {y^4} + {x^2} - 2x{y^2} = 2y \\
   \Rightarrow {y^4} + {x^2} - 2x{y^2} - 2y = 2y - 2y \\
   \Rightarrow {y^4} + {x^2} - 2x{y^2} - 2y = 0 \\
 $
Since, we obtained a sorted, so differentiating both the sides of the above equation with respect to $x$ :
\[
  {y^4} + {x^2} - 2x{y^2} - 2y = 0 \\
   \Rightarrow \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = \dfrac{{d0}}{{dx}} \\
   \Rightarrow \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = 0 \\
 \]
Splitting the terms for differentiation:
\[
  \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = 0 \\
   \Rightarrow \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - \dfrac{{d\left( {2x{y^2}} \right)}}{{dx}} - \dfrac{{d\left( {2y} \right)}}{{dx}} = 0 \\
 \]
Taking out the constants from the parenthesis:
\[
  \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - \dfrac{{d\left( {2x{y^2}} \right)}}{{dx}} - \dfrac{{d\left( {2y} \right)}}{{dx}} = 0 \\
   \Rightarrow \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - 2\dfrac{{d\left( {x{y^2}} \right)}}{{dx}} - 2\dfrac{{d\left( y \right)}}{{dx}} = 0 \\
 \] …….(A)
We would solve each operand separately:
Starting with \[\dfrac{{d{y^4}}}{{dx}}\].
Since, we know that \[\dfrac{{d{y^n}}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}\].
So, from this we get: \[\dfrac{{d{y^4}}}{{dx}} = 4{y^{4 - 1}}\dfrac{{dy}}{{dx}} = 4{y^3}\dfrac{{dy}}{{dx}}\]. ………………..(1)
Then we have \[\dfrac{{d{x^2}}}{{dx}}\] :
Since, we know that \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\].
So, from this we get: \[\dfrac{{d{x^2}}}{{dx}} = 2{x^{2 - 1}} = 2x\]. ………………..(2)
Then we have \[\dfrac{{d\left( {x{y^2}} \right)}}{{dx}}\] :
Since, we know that it is in product form, and it can differentiated as \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + v\dfrac{{d\left( u \right)}}{{dx}}\].
Comparing \[\left( {uv} \right)\] with \[\left( {x{y^2}} \right)\], we get \[u = x,v = {y^2}\]
So, differentiating it from the product rule, we get:
\[
  \dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = x\dfrac{{d\left( {{y^2}} \right)}}{{dx}} + {y^2}\dfrac{{d\left( x \right)}}{{dx}} \\
   \Rightarrow \dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = x.2y\dfrac{{dy}}{{dx}} + {y^2}\left( 1 \right) \\
   \Rightarrow \dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = 2xy\dfrac{{dy}}{{dx}} + {y^2} \\
 \]. ………………..(3)
Then we have \[\dfrac{{dy}}{{dx}}\] :
Since, we can see that it cannot be further simplified.
Therefore, it is \[\dfrac{{dy}}{{dx}}\] only ………………..(4)
Substituting (1), (2), (3) and (4) in A, we get:
\[
  \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - 2\dfrac{{d\left( {x{y^2}} \right)}}{{dx}} - 2\dfrac{{d\left( y \right)}}{{dx}} = 0 \\
   \Rightarrow 4{y^3}\dfrac{{dy}}{{dx}} + 2x - 2\left( {2xy\dfrac{{dy}}{{dx}} + {y^2}} \right) - 2\dfrac{{dy}}{{dx}} = 0 \\
 \]
Opening the parenthesis above, and we get:
\[
  4{y^3}\dfrac{{dy}}{{dx}} + 2x - 2\left( {2xy\dfrac{{dy}}{{dx}} + {y^2}} \right) - 2\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow 4{y^3}\dfrac{{dy}}{{dx}} + 2x - 4xy\dfrac{{dy}}{{dx}} - 2{y^2} - 2\dfrac{{dy}}{{dx}} = 0 \\
 \]
Taking \[2\] common:
\[2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right) = 0\]
Dividing both the sides by \[2\]:
\[
  2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right) = 0 \\
   \Rightarrow \dfrac{{2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right)}}{2} = 0 \\
   \Rightarrow 2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}} = 0 \\
 \]
Taking coefficient of \[\dfrac{{dy}}{{dx}}\] in one parenthesis:
\[
  2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} = 0 \\
 \]
Adding both the sides by \[{y^2}\] and subtracting \[x\] , we get:
\[
  \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} = 0 \\
   \Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} + {y^2} - x = {y^2} - x \\
   \Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} = {y^2} - x \\
 \]
Dividing both the sides by \[\left( {2{y^3} - 2xy - 1} \right)\] , and we get:
\[
  \dfrac{{\left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}}}}{{\left( {2{y^3} - 2xy - 1} \right)}} = \dfrac{{{y^2} - x}}{{\left( {2{y^3} - 2xy - 1} \right)}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{\left( {2{y^3} - 2xy - 1} \right)}} \\
 \]
Which is proved.
Therefore, if $y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } $ , then $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{2{y^3} - 2xy - 1}}$.
Hence, proved.

Note:
1. Always preferred to go step by step for ease, otherwise there is a huge chance of mistakes to arise.
2. It’s important to square the terms to remove the roots, otherwise, it would become much more complicated to differentiate.