d dx x x = - Vidyadip

MCQ

${d \over {dx}}\sqrt {x\sin x} = $

✓
${{\sin x + x\cos x} \over {2\sqrt {x\sin x} }}$
B
${{\sin x + x\cos x} \over {\sqrt {x\sin x} }}$
C
${{x\sin x + \cos x} \over {\sqrt {2\sin x} }}$
D
${{\sin x + x\cos x} \over {\sqrt {2x\sin x} }}$

✓

Answer

Correct option: A.

${{\sin x + x\cos x} \over {2\sqrt {x\sin x} }}$

a
(a) Let ${y^2} = x\sin x \Rightarrow 2y\frac{{dy}}{{dx}} = \sin x + x\cos x$

$\therefore \frac{{dy}}{{dx}} = \frac{{[\sin x + x\cos x]}}{{2\sqrt {x\sin x} }}$.

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