MCQ
જો $y = {x^2}\log x + {2 \over {\sqrt x }},$ તો ${{dy} \over {dx}} = $
- A$x + 2x\log x - {1 \over {\sqrt x }}$
- ✓$x + 2x\log x - {1 \over {{x^{3/2}}}}$
- C$x + 2x\log x - {2 \over {{x^{3/2}}}}$
- Dએકપણ નહીં
$\frac{{dy}}{{dx}} = 2x\log x + x - {x^{ - 3/2}} = x + 2x\log x - \frac{1}{{{x^{3/2}}}}$.
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