Derivative Product 3B219F

1. **State the problem:** Find the derivative of the function $$f(t) = e^{5t} \sin(2t)$$. 2. **Formula used:** We will use the product rule for derivatives, which states: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$ where $$u(t) = e^{5t}$$ and $$v(t) = \sin(2t)$$. 3. **Find derivatives of each part:** - Derivative of $$u(t) = e^{5t}$$ is $$u'(t) = 5e^{5t}$$ by chain rule. - Derivative of $$v(t) = \sin(2t)$$ is $$v'(t) = 2\cos(2t)$$ by chain rule. 4. **Apply product rule:** $$\frac{d}{dt}[e^{5t} \sin(2t)] = 5e^{5t} \sin(2t) + e^{5t} \cdot 2\cos(2t)$$ 5. **Simplify the expression:** $$= e^{5t} (5 \sin(2t) + 2 \cos(2t))$$ **Final answer:** $$\boxed{\frac{d}{dt} \left(e^{5t} \sin(2t)\right) = e^{5t} (5 \sin(2t) + 2 \cos(2t))}$$