If `y=f((2x-1)/(x^2+1))` and `f^(prime)(x)=sinx^2` , find `(dy)/(dx)` .

Explanation

\begin{equation}
\begin{aligned}
&y=f\left(\frac{2 x-1}{x^{2}+1}\right) \\
&f^{\prime}(x)=\sin x^{2} \\
&\frac{d y}{d x}=f^{\prime}\left(\frac{2 x-1}{x^{2}+1}\right) \frac{d}{d x}\left(\frac{2 x-1}{x^{2}+1}\right) \\
&=\sin \left(\frac{2 x_{1}}{x^{2}+1}\right)^{2}\left(\frac{2\left(x^{2}+1\right)-2 x(2 x-1)}{\left(x^{2}+1\right)^{2}}\right) \\
&=\frac{2+2 x-2 x^{2}}{\left(x^{2}+1\right)^{2}} \sin \left(\frac{2 x-1}{x^{2}+1}\right)^{2}
\end{aligned}
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