If \[y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]and \[{f}'(x)=\sin {{x}^{2}},\]then \[\frac{dy}{dx}=\] [IIT 1982]

\[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\]

\[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]

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

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

If \[y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]and \[{f}'(x)=\sin {{

1.	If \[y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]and \[{f}'(x)=\sin {{x}^{2}},\]then \[\frac{dy}{dx}=\] [IIT 1982]
A.	\[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\]
B.	\[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]
C.	\[\frac{-2{{x}^{2}}+2x+2}{{{({{x}^{2}}+1)}^{2}}}{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]
D.	\[\frac{-2{{x}^{2}}+2x+2}{{{({{x}^{2}}+1)}^{2}}}\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\]
Answer» E.

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If \[y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]and \[{f}'(x)=\sin {{x}^{2}},\]then \[\frac{dy}{dx}=\] [IIT 1982]

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