JEE Advance - Mathematics (1982 - No. 35)

If $$y = f\left( {{{2x - 1} \over {{x^2} + 1}}} \right)$$ and $$f'\left( x \right) = \sin {x^2}$$, then $${{dy} \over {dx}} = ..........$$

$\frac{2 + 2x - 2x^2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$

$\frac{2 - 2x + 2x^2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$

$\frac{2 + 2x - 2x^2}{(x^2 + 1)^2} \cos \left( \frac{2x - 1}{x^2 + 1} \right)^2$

$\frac{2 - 2x - 2x^2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$

$\frac{2 + 2x + 2x^2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$