If f(x) = int_0^x {tsin t,dt} ,then f'(x) = | vedclass

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(B) Given that $f(x) = \int_0^x {t\sin t\,dt}$.According to the Leibniz integral rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $f'(x) = F(h(x

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$x\sin x$

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(B) Given that $f(x) = \int_0^x {t\sin t\,dt}$.According to the Leibniz integral rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $f'(x) = F(h(x), x) \cdot h'(x) - F(g(x), x) \cdot g'(x) + \int_{g(x)}^{h(x)} \frac{\partial}{\partial x} F(t, x) \, dt$.In this case,$F(t, x) = t\sin t$,$h(x) = x$,and $g(x) = 0$.Since the integrand $t\sin t$ does not depend on $x$,the partial derivative $\frac{\partial}{\partial x} (t\sin t) = 0$.Therefore,$f'(x) = (x\sin x) \cdot \frac{d}{dx}(x) - (0\sin 0) \cdot \frac{d}{dx}(0) + 0$.$f'(x) = x\sin x \cdot 1 - 0 = x\sin x$.

If $f(x) = \int_0^x {t\sin t\,dt} $,then $f'(x) = $

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