Explanation
$$y=\sqrt { x+\sqrt { x+\sqrt { x+......\infty } } } \\ \therefore y=\sqrt { x+y } \\ Squaring\quad on\quad both\ sides,\\ y^{ 2 }=x+y\\ Differentiating\ on\ both\ sides,\\ 2y\dfrac { dy }{ dx }=1+\dfrac { dy }{ dx } \\ \therefore \dfrac { dy }{ dx }=\dfrac { 1 }{ 2y-1 } \\ \Rightarrow \dfrac { 1 }{ \sqrt { (2y-1)^{ 2 } } }=\dfrac { 1 }{ \sqrt { 4y^{ 2 }-4y+1 } }=\dfrac { 1 }{ \sqrt { 1+4x } } \\ \Rightarrow \dfrac { \sqrt { x+y } }{ \sqrt { (1 + 4x)(x+y) } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + x + y } } \\\Rightarrow \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + y^{ 2 } } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { (2x+y)^{ 2 } } } = \dfrac { y }{ 2x + y }$$
$$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right) $$ and $$ \displaystyle g^{ ' }\left( x \right) =-f\left( x \right)$$ for all real x and $$ \displaystyle f\left( 5 \right) =2=f^{ ' }\left( 5 \right) $$ then $$ \displaystyle f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ is -
Given, $$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right)$$ and $$g^{ ' }\left( x \right) =-f\left( x \right)$$
Now $$ \displaystyle \frac { d }{ dx } \left[ f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right) \right] =2f\left( x \right) f^{ ' }\left( x \right) +2g\left( x \right) g^{ ' }\left( x \right) $$
$$ \displaystyle =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right)$$ =constant
$$ \displaystyle f^{ 2 }\left( 5 \right) +g^{ 2 }\left( 5 \right) = 4 + 4 = 8$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ = 8
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