1. Vamos derivar a função dada: $$y = \arctan\left(\frac{1 + \cos(5x)}{1 - \cos(5x)}\right)$$. 2. Para derivar uma função do tipo $$y = \arctan(u)$$, usamos a regra: $$\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx}$$. 3. Definimos $$u = \frac{1 + \cos(5x)}{1 - \cos(5x)}$$. Agora, vamos calcular $$\frac{du}{dx}$$ usando a regra do quociente: $$\frac{du}{dx} = \frac{(1 - \cos(5x)) \cdot \frac{d}{dx}(1 + \cos(5x)) - (1 + \cos(5x)) \cdot \frac{d}{dx}(1 - \cos(5x))}{(1 - \cos(5x))^2}$$. 4. Calculando as derivadas: $$\frac{d}{dx}(1 + \cos(5x)) = -5 \sin(5x)$$ $$\frac{d}{dx}(1 - \cos(5x)) = 5 \sin(5x)$$. 5. Substituindo na expressão de $$\frac{du}{dx}$$: $$\frac{du}{dx} = \frac{(1 - \cos(5x))(-5 \sin(5x)) - (1 + \cos(5x))(5 \sin(5x))}{(1 - \cos(5x))^2}$$. 6. Simplificando o numerador: $$-5 \sin(5x)(1 - \cos(5x)) - 5 \sin(5x)(1 + \cos(5x)) = -5 \sin(5x)(1 - \cos(5x) + 1 + \cos(5x)) = -5 \sin(5x)(2) = -10 \sin(5x)$$. 7. Portanto: $$\frac{du}{dx} = \frac{-10 \sin(5x)}{(1 - \cos(5x))^2}$$. 8. Agora, calculamos $$1 + u^2$$: $$1 + \left(\frac{1 + \cos(5x)}{1 - \cos(5x)}\right)^2 = \frac{(1 - \cos(5x))^2 + (1 + \cos(5x))^2}{(1 - \cos(5x))^2}$$. 9. Expandindo os quadrados no numerador: $$(1 - \cos(5x))^2 + (1 + \cos(5x))^2 = (1 - 2\cos(5x) + \cos^2(5x)) + (1 + 2\cos(5x) + \cos^2(5x)) = 2 + 2\cos^2(5x)$$. 10. Assim: $$1 + u^2 = \frac{2 + 2\cos^2(5x)}{(1 - \cos(5x))^2} = \frac{2(1 + \cos^2(5x))}{(1 - \cos(5x))^2}$$. 11. Finalmente, a derivada de $$y$$ é: $$\frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} = \frac{1}{\frac{2(1 + \cos^2(5x))}{(1 - \cos(5x))^2}} \cdot \frac{-10 \sin(5x)}{(1 - \cos(5x))^2}$$. 12. Simplificando: $$\frac{dy}{dx} = \frac{(1 - \cos(5x))^2}{2(1 + \cos^2(5x))} \cdot \frac{-10 \sin(5x)}{(1 - \cos(5x))^2} = \frac{-10 \sin(5x)}{2(1 + \cos^2(5x))} = \frac{-5 \sin(5x)}{1 + \cos^2(5x)}$$. Resposta final: $$\boxed{\frac{dy}{dx} = \frac{-5 \sin(5x)}{1 + \cos^2(5x)}}$$