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
If $$ \displaystyle x\sqrt { \left( 1+y \right) } +y\sqrt { \left( 1+x \right) } =0$$. then $$ \displaystyle \frac{dy}{dx}$$ equals -
$$ \displaystyle x\sqrt { \left( 1+y \right) } +y\sqrt { \left( 1+x \right) } =0$$.
$$\Rightarrow \displaystyle x\sqrt { \left( 1+y \right) } =-y\sqrt { \left( 1+x \right) } $$.
Squaring both sides, $$ \displaystyle { x }^{ 2 }\left( 1+y \right)={ y }^{ 2 }\left( 1+x \right) \Rightarrow { x }^{ 2 }-{ y }^{ 2 }+xy\left( x-y \right) = 0$$
$$ \displaystyle \Rightarrow \left( x-y \right) \left( x+y+xy \right) = 0$$
Now $$ \displaystyle x\neq y$$ [does not satisfy the given equation]
$$ \displaystyle \therefore x+y+xy=0 \Rightarrow y=\frac { -x }{ 1+x } $$
$$ \displaystyle \therefore \quad \frac { dy }{ dx } =\frac { -\left( 1+x \right) +x }{ \left( 1+x \right) ^{ 2 } } =-\frac { 1 }{ \left( 1+x \right) ^{ 2 } } $$
If $$ \displaystyle y=\dfrac { x }{ a+\dfrac { x }{ b+y } } $$, then $$ \displaystyle \frac{dy}{dx}$$ is
$$\displaystyle y=\dfrac { x }{ a+\dfrac { x }{ b+y } }$$
$$\displaystyle y=\frac { x(b+y) }{ ab+ay+x }$$
$$ \displaystyle \Rightarrow $$ $$aby+ay^{2}+xy = xb+xy$$
$$ \displaystyle \Rightarrow $$ $$aby+ay^{2} = xb$$
Differentiating both side w.r.t $$x$$ we get,
$$ \displaystyle \Rightarrow ab\frac { dy }{ dx } +2ay\frac { dy }{ dx }=b \Rightarrow \frac { dy }{ dx } =\frac { b }{ ab+2ay }$$
Find $$ \displaystyle \frac { dy }{ dx } $$, if $$x + y =\displaystyle \sin { \left( x-y \right) } $$
Please disable the adBlock and continue. Thank you.