If $f\left( x \right) = \int\limits_0^x {t\sin \,t\,dt}$ ,then $f'\left( x \right)$ is

Option b is correct

Let$f\left( x \right) = \int\limits_0^x {t\sin t} dt$

$\Rightarrow$ $f\left( x \right) = \left[ {t\left( { - \cos \,t} \right) - \int {1 \cdot \left( { - \cos \,t} \right)dt} } \right]_0^x = \left[ { - t\cos t + \sin t} \right]_0^x$
$\Rightarrow$ $f\left( x \right) = - x\cos x + \sin x$

$\therefore$ $f'\left( x \right) = - \left[ {x\left( { - \sin x} \right) + \cos x} \right] + \cos x = x\sin x - \cos x + \cos x$
$= x\sin x$

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