Derivative Product 33F38F

1. **State the problem:** We need to find the derivative of the function $$F(t) = e^{5t} \sin(2t)$$ with respect to $$t$$. 2. **Formula used:** Since $$F(t)$$ is a product of two functions, $$u(t) = e^{5t}$$ and $$v(t) = \sin(2t)$$, we use the product rule: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$ 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:** $$F'(t) = u'(t)v(t) + u(t)v'(t) = 5e^{5t} \sin(2t) + e^{5t} \cdot 2\cos(2t)$$ 5. **Factor common terms:** $$F'(t) = e^{5t} (5 \sin(2t) + 2 \cos(2t))$$ 6. **Final answer:** $$\boxed{F'(t) = e^{5t} (5 \sin(2t) + 2 \cos(2t))}$$