If y = x (2x + 3) ^2 x + 1 , then dy dx = - Vidyadip

MCQ

If $y = {{\sqrt x {{(2x + 3)}^2}} \over {\sqrt {x + 1} }},$ then ${{dy} \over {dx}} = $

✓
$y{\rm{ }}\left[ {{1 \over {2x}} + {4 \over {2x + 3}} - {1 \over {2(x + 1)}}} \right]$
B
$y{\rm{ }}\left[ {{1 \over {3x}} + {4 \over {2x + 3}} + {1 \over {2(x + 1)}}} \right]$
C
$y{\rm{ }}\left[ {{1 \over {3x}} + {4 \over {2x + 3}} + {1 \over {x + 1}}} \right]$
D
None of these

✓

Answer

Correct option: A.

$y{\rm{ }}\left[ {{1 \over {2x}} + {4 \over {2x + 3}} - {1 \over {2(x + 1)}}} \right]$

a
(a) $y = \frac{{\sqrt x {{(2x + 3)}^2}}}{{\sqrt {x + 1} }} $

$\Rightarrow \log y = \frac{1}{2}\log x + 2\log (2x + 3) - \frac{1}{2}\log (x + 1)$

$ \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2x}} + \frac{{2.2}}{{(2x + 3)}} - \frac{1}{{2(x + 1)}}$

or $\frac{{dy}}{{dx}} = y\left[ {\frac{1}{{2x}} + \frac{4}{{2x + 3}} - \frac{1}{{2(x + 1)}}} \right]$.

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