1. **Problem:** Find the derivative $\frac{dy}{dx}$ of the function $$y = e^{\cos x} \times \sin^3(x)$$ 2. **Formula and rules:** - Use the product rule: $$\frac{d}{dx}[u \cdot v] = u'v + uv'$$ - Use the chain rule for composite functions. 3. **Identify parts:** - Let $u = e^{\cos x}$ and $v = \sin^3(x)$. 4. **Find derivatives:** - $$u' = e^{\cos x} \cdot \frac{d}{dx}(\cos x) = e^{\cos x} \cdot (-\sin x) = -\sin x e^{\cos x}$$ - $$v = (\sin x)^3$$ so by chain rule, $$v' = 3 \sin^2(x) \cdot \cos x$$ 5. **Apply product rule:** $$\frac{dy}{dx} = u'v + uv' = (-\sin x e^{\cos x}) \cdot \sin^3(x) + e^{\cos x} \cdot 3 \sin^2(x) \cos x$$ 6. **Simplify:** $$\frac{dy}{dx} = -\sin x \sin^3(x) e^{\cos x} + 3 \sin^2(x) \cos x e^{\cos x} = e^{\cos x} \sin^2(x) (-\sin^2 x + 3 \cos x)$$ **Final answer:** $$\boxed{\frac{dy}{dx} = e^{\cos x} \sin^2(x) (3 \cos x - \sin^2 x)}$$