If $y=f\left(\frac{2 x-1}{x^2+1}\right)$ and $f^{\prime}(x)=\sin x^2$, then $\frac{d y}{d x}$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$y=f\left(\frac{2 x-1}{x^2+1}\right)$
$\Rightarrow \frac{d y}{d x} =f'\left(\frac{2 x-1}{x^2+1}\right) \frac{d}{d x}\left(\frac{2 x-1}{x^2+1}\right)$
$\Rightarrow \frac{d y}{d x}=f'\left(\frac{2 x-1}{x^2+1}\right)\left\{\frac{2\left(x^2+1\right)-2 x(2 x-1)}{\left(x^2+1\right)^2}\right\}$ $\left[∵ f'(x)=\sin x^2\right]$
$\Rightarrow \frac{d y}{d x}=\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left\{\frac{2+2 x-2 x^2}{\left(x^2+1\right)^2}\right\}$