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

Last updated date: 13th Jul 2024
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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.