If $t\left( 1+{ x }^{ 2 } \right) =x$ and ${ x }^{ 2 }+{ t }^{ 2 }=y,$ then at $x=2,$ the value of $\displaystyle\frac{dy}{dx}$

$\displaystyle { x }^{ 2 }+{ t }^{ 2 }=y\Rightarrow \frac { dy }{ dx } =2x+2t.\frac { dt }{ dx }$ ...(1)

As $\displaystyle t=\frac { x }{ 1+{ x }^{ 2 } } \Rightarrow \frac { dt }{ dx } =\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

Substitute these value of $t$ and $\displaystyle\frac{dt}{dx}$ in (1), we get

$\displaystyle \frac { dy }{ dx } =2x+\frac { 2x }{ 1+{ x }^{ 2 } } .\frac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } }$

On ptiing $x=2$ in $\displaystyle\frac{dy}{dx}$, we get

$\displaystyle \frac { dy }{ dx } =\frac { 488 }{ 125 }$