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

A
$\frac{24}{5}$
B
$\frac{101}{125}$
C
$\frac{488}{125}$
D
$\frac{358}{125}$

Solution:

$Given t = \frac{x}{1 + x^{2}} \Rightarrow \frac{d t}{d x} = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}}; \Rightarrow x^{2} + t^{2} = y$
$\frac{d y}{d x} = 2 x + 2 t \frac{d t}{d x} = 2 x + 2 \frac{x}{1 + x^{2}} \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \Rightarrow x = 2 \Rightarrow \frac{d y}{d x} = 4 + \frac{4}{5} \frac{(- 3)}{25} = \frac{488}{125}$

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

Solution:

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