Derivatives Exponential Trig 46504F

1. **State the problem:** Find the derivatives of the functions $$F(t) = e^{5t} \sin(2t)$$ and $$y = \sqrt[3]{e^x + 2} = (e^x + 2)^{\frac{1}{3}}.$$\n\n2. **Derivative of $F(t)$:** Use the product rule: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$ where $$u(t) = e^{5t}$$ and $$v(t) = \sin(2t).$$\n\n3. Compute derivatives of each part:\n$$u'(t) = \frac{d}{dt} e^{5t} = 5 e^{5t}$$\n$$v'(t) = \frac{d}{dt} \sin(2t) = 2 \cos(2t)$$\n\n4. Apply product rule:\n$$F'(t) = u'(t)v(t) + u(t)v'(t) = 5 e^{5t} \sin(2t) + e^{5t} (2 \cos(2t)) = e^{5t} (5 \sin(2t) + 2 \cos(2t))$$\n\n5. **Derivative of $y$:** Rewrite $$y = (e^x + 2)^{\frac{1}{3}}.$$ Use the chain rule: $$\frac{dy}{dx} = \frac{1}{3} (e^x + 2)^{-\frac{2}{3}} \cdot \frac{d}{dx}(e^x + 2).$$\n\n6. Compute inner derivative:\n$$\frac{d}{dx}(e^x + 2) = e^x$$\n\n7. Substitute back:\n$$\frac{dy}{dx} = \frac{1}{3} (e^x + 2)^{-\frac{2}{3}} e^x = \frac{e^x}{3 (e^x + 2)^{\frac{2}{3}}}.$$\n\n**Final answers:**\n$$F'(t) = e^{5t} (5 \sin(2t) + 2 \cos(2t))$$\n$$\frac{dy}{dx} = \frac{e^x}{3 (e^x + 2)^{\frac{2}{3}}}$$