1. **State the problem:** We have the curve defined by $$y = \frac{e^{\sin x}}{\cos^2 x}$$ for $$0 \leq x \leq 2\pi$$. We need to find $$\frac{dy}{dx}$$ and then find the $$x$$-coordinates of the stationary points where $$\frac{dy}{dx} = 0$$. 2. **Differentiate using the quotient rule:** Let $$u = e^{\sin x}$$ and $$v = \cos^2 x$$. The quotient rule states: $$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ 3. **Find $$\frac{du}{dx}$$:** $$u = e^{\sin x}$$ Using chain rule: $$\frac{du}{dx} = e^{\sin x} \cdot \cos x$$ 4. **Find $$\frac{dv}{dx}$$:** $$v = \cos^2 x = (\cos x)^2$$ Using chain rule: $$\frac{dv}{dx} = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x$$ 5. **Substitute into quotient rule:** $$\frac{dy}{dx} = \frac{\cos^2 x \cdot e^{\sin x} \cos x - e^{\sin x} \cdot (-2 \cos x \sin x)}{\cos^4 x}$$ Simplify numerator: $$e^{\sin x} \cos x \left( \cos^2 x + 2 \sin x \right)$$ So: $$\frac{dy}{dx} = \frac{e^{\sin x} \cos x (\cos^2 x + 2 \sin x)}{\cos^4 x} = e^{\sin x} \frac{\cos^2 x + 2 \sin x}{\cos^3 x}$$ 6. **Find stationary points:** Set $$\frac{dy}{dx} = 0$$. Since $$e^{\sin x} \neq 0$$ and $$\cos^3 x \neq 0$$ except where undefined, the numerator must be zero: $$\cos^2 x + 2 \sin x = 0$$ Rewrite: $$\cos^2 x = -2 \sin x$$ Using $$\cos^2 x = 1 - \sin^2 x$$: $$1 - \sin^2 x = -2 \sin x$$ Rearranged: $$\sin^2 x - 2 \sin x - 1 = 0$$ 7. **Solve quadratic in $$\sin x$$:** Let $$s = \sin x$$: $$s^2 - 2s - 1 = 0$$ Use quadratic formula: $$s = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$$ 8. **Check valid solutions:** $$1 + \sqrt{2} > 1$$ (not possible for sine) $$1 - \sqrt{2} \approx -0.414$$ (valid) So: $$\sin x = 1 - \sqrt{2}$$ 9. **Find $$x$$ in $$[0, 2\pi]$$:** $$x = \arcsin(1 - \sqrt{2})$$ and $$x = \pi - \arcsin(1 - \sqrt{2})$$ **Final answers:** $$\frac{dy}{dx} = e^{\sin x} \frac{\cos^2 x + 2 \sin x}{\cos^3 x}$$ Stationary points at $$x = \arcsin(1 - \sqrt{2})$$ and $$x = \pi - \arcsin(1 - \sqrt{2})$$ within $$0 \leq x \leq 2\pi$$.