If g(x) = int_{sin x}^{sin(2x)} sin^{-1}(t) , | vedclass

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(C) Given $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$. Using Leibniz's Rule for differentiation under the integral sign: $g^{\prime}(x) = \si

Vedclass · Accepted Answer

$g^{\prime}\left(\frac{\pi}{2}\right) = 0$

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(C) Given $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$. Using Leibniz's Rule for differentiation under the integral sign: $g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot \frac{d}{dx}(\sin(2x)) - \sin^{-1}(\sin x) \cdot \frac{d}{dx}(\sin x)$. $g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot (2\cos(2x)) - \sin^{-1}(\sin x) \cdot (\cos x)$. For $x = \frac{\pi}{2}$: $g^{\prime}\left(\frac{\pi}{2}\right) = \sin^{-1}(\sin(\pi)) \cdot (2\cos(\pi)) - \sin^{-1}(\sin(\frac{\pi}{2})) \cdot (\cos(\frac{\pi}{2}))$. Since $\sin(\pi) = 0$,$\sin^{-1}(0) = 0$. Since $\cos(\frac{\pi}{2}) = 0$,the second term is also $0$. Thus,$g^{\prime}\left(\frac{\pi}{2}\right) = 0 \cdot (-2) - (\frac{\pi}{2}) \cdot 0 = 0$. Therefore,option $C$ is correct.

If $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$,then

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