If y = x^2 x + 2 x , then dy dx = - Vidyadip

MCQ

If $y = {x^2}\log x + {2 \over {\sqrt x }},$ then ${{dy} \over {dx}} = $

A
$x + 2x\log x - {1 \over {\sqrt x }}$
✓
$x + 2x\log x - {1 \over {{x^{3/2}}}}$
C
$x + 2x\log x - {2 \over {{x^{3/2}}}}$
D
None of these

✓

Answer

Correct option: B.

$x + 2x\log x - {1 \over {{x^{3/2}}}}$

b
(b) $y = {x^2}\log x + \frac{2}{{\sqrt x }}$

$\frac{{dy}}{{dx}} = 2x\log x + x - {x^{ - 3/2}} = x + 2x\log x - \frac{1}{{{x^{3/2}}}}$.

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