If \[y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}},\]then \[\frac{dy}{dx}=\] [AISSE 1986]

\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)} \right]\]

\[y\text{ }\left[ \frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)} \right]\]

\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{x+1} \right]\]

If \[y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}},\]then \[\frac{dy}{dx

1.	If \[y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}},\]then \[\frac{dy}{dx}=\] [AISSE 1986]
A.	\[y\text{ }\left[ \frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)} \right]\]
B.	\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)} \right]\]
C.	\[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{x+1} \right]\]
D.	None of these
Answer» B. \[y\text{ }\left[ \frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)} \right]\]

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If \[y=\frac{\sqrt{x}{{(2x+3)}^{2}}}{\sqrt{x+1}},\]then \[\frac{dy}{dx}=\] [AISSE 1986]

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