If y = 5x [3] (1 - x) ^2 + ^2 (2x + 1), then dy dx = - Vidyadip

MCQ

If $y = \frac{{5x}}{{\sqrt[3]{{{{(1 - x)}^2}}}}} + {\cos ^2}(2x + 1)$, then $\frac{{dy}}{{dx}} = $

✓
$\frac{{5(3 - x)}}{{3{{(1 - x)}^{5/3}}}} - 2\sin (4x + 2)$
B
$\frac{{5(3 - x)}}{{3{{(1 - x)}^{2/3}}}} - 2\sin (4x + 4)$
C
$\frac{{5(3 - x)}}{{3{{(1 - x)}^{2/3}}}} - 2\sin (2x + 1)$
D
None of these

✓

Answer

Correct option: A.

$\frac{{5(3 - x)}}{{3{{(1 - x)}^{5/3}}}} - 2\sin (4x + 2)$

a
(a) $y = 5x{(1 - x)^{ - 2/3}} + {\cos ^2}(2x + 1)$
$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{10x}}{{3{{(1 - x)}^{5/3}}}} + \frac{5}{{{{(1 - x)}^{2/3}}}} - 4\cos (2x + 1)\sin (2x + 1)$
$ = \frac{5}{{{{(1 - x)}^{2/3}}}}\left[ {\frac{{2x}}{{3(1 - x)}} + 1} \right] - 2\sin (4x + 2)$
$ = \frac{{5(3 - x)}}{{3{{(1 - x)}^{5/3}}}} - 2\sin (4x + 2)$.

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