If \[y={{\tan }^{-1}}\frac{4x}{1+5{{x}^{2}}}+{{\tan }^{-1}}\frac{2+3x}{3-2x}\], then \[\frac{dy}{dx}=\]

\[\frac{1}{1+25{{x}^{2}}}+\frac{2}{1+{{x}^{2}}}\]

\[\frac{5}{1+25{{x}^{2}}}+\frac{2}{1+{{x}^{2}}}\]

If \[y={{\tan }^{-1}}\frac{4x}{1+5{{x}^{2}}}+{{\tan }^{-1}}\frac{2+3x}

1.	If \[y={{\tan }^{-1}}\frac{4x}{1+5{{x}^{2}}}+{{\tan }^{-1}}\frac{2+3x}{3-2x}\], then \[\frac{dy}{dx}=\]
A.	\[\frac{1}{1+25{{x}^{2}}}+\frac{2}{1+{{x}^{2}}}\]
B.	\[\frac{5}{1+25{{x}^{2}}}+\frac{2}{1+{{x}^{2}}}\]
C.	\[\frac{5}{1+25{{x}^{2}}}\]
D.	\[\frac{1}{1+25{{x}^{2}}}\]
Answer» D. \[\frac{1}{1+25{{x}^{2}}}\]

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If \[y={{\tan }^{-1}}\frac{4x}{1+5{{x}^{2}}}+{{\tan }^{-1}}\frac{2+3x}{3-2x}\], then \[\frac{dy}{dx}=\]

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