What is the differential of $$\frac{d}{dt}(\sin 4t u(t-3))$$? Multiple Choice $$4 \cos 4t u(t-3) - 0.5366 \delta(t-3)$$ $$0.5366 u(t-3) - 4 \cos 4t \delta(t-3)$$ $$0.5366 u(t-3) + 4 \cos 4t \delta(t-3)$$ $$4 \cos 4t u(t-3) + 0.5366 \delta(t-3)$$
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The product rule for differentiation is given by: $$\frac{d}{dt}(f(t)g(t)) = f'(t)g(t) + f(t)g'(t)$$ Show more…
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