Slope of the tangent to the curve y=2 e^x sin | vedclass

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(B) $y = 2 e^x \sin \left(\frac{\pi}{4} - \frac{x}{2}ight) \cos \left(\frac{\pi}{4} - \frac{x}{2}ight)$ Using the identity $2 \sin 	heta \cos 	h

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$\pi$

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(B) $y = 2 e^x \sin \left(\frac{\pi}{4} - \frac{x}{2}ight) \cos \left(\frac{\pi}{4} - \frac{x}{2}ight)$ Using the identity $2 \sin 	heta \cos 	heta = \sin 2 	heta$,we get: $y = e^x \sin \left(2 \left(\frac{\pi}{4} - \frac{x}{2}ight)ight) = e^x \sin \left(\frac{\pi}{2} - xight) = e^x \cos x$ The slope of the tangent is given by $\frac{dy}{dx}$: $\frac{dy}{dx} = e^x \cos x + e^x (-\sin x) = e^x (\cos x - \sin x)$ Let $T(x) = e^x (\cos x - \sin x)$. To find the minimum,we find $\frac{dT}{dx}$: $\frac{dT}{dx} = e^x (\cos x - \sin x) + e^x (-\sin x - \cos x) = e^x (\cos x - \sin x - \sin x - \cos x) = -2 e^x \sin x$ Setting $\frac{dT}{dx} = 0$ for critical points: $-2 e^x \sin x = 0 \implies \sin x = 0$ In the interval $0 \leq x \leq 2 \pi$,$x = 0, \pi, 2 \pi$. Evaluating $T(x)$ at these points: $T(0) = e^0 (\cos 0 - \sin 0) = 1$ $T(\pi) = e^\pi (\cos \pi - \sin \pi) = -e^\pi$ $T(2 \pi) = e^{2 \pi} (\cos 2 \pi - \sin 2 \pi) = e^{2 \pi}$ Comparing the values,the minimum value is $-e^\pi$ at $x = \pi$.

Slope of the tangent to the curve $y=2 e^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)$ where $0 \leq x \leq 2 \pi$ is minimum at $x=$

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