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If $t (1 + x^{2}) = x and x^{2} + t^{2} = y then at x = 2, \frac{d y}{d x} =$

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\frac{24}{5}

\frac{101}{125}

\frac{488}{125}

\frac{358}{125}

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detailed solution

Correct option is C

$Given t = \frac{x}{1 + x^{2}} \Rightarrow \frac{d t}{d x} = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}};$

$\Rightarrow x^{2} + t^{2} = y$
$\frac{d y}{d x} = 2 x + 2 t \frac{d t}{d x} = 2 x + 2 \frac{x}{1 + x^{2}} \frac{1 - x^{2}}{{(1 + x^{2})}^{2}}$

$\Rightarrow x = 2 \Rightarrow \frac{d y}{d x} = 4 + \frac{4}{5} \frac{(- 3)}{25} = \frac{488}{125}$

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If t1+x2=x and x2+t2=y then at x=2,dydx=

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If $t (1 + x^{2}) = x and x^{2} + t^{2} = y then at x = 2, \frac{d y}{d x} =$