Given: \[ y = \sin(\sin(\sin(2x))) \] We are to find \(\frac{dy}{dx}\). Let's proceed step by step using the **chain rule**. --- Let: - \(u = 2x\) - \(v = \sin(u)\) - \(w = \sin(v)\) - \(y = \sin(w)\) So, \[ y = \sin(w), \quad w = \sin(v), \quad v = \sin(u), \quad u = 2x \] By the chain rule: \[ \frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} \] Now, compute each derivative: - \(\frac{dy}{dw} = \cos(w)\) - \(\frac{dw}{dv} = \cos(v)\) - \(\frac{dv}{du} = \cos(u)\) - \(\frac{du}{dx} = 2\) Substitute back the expressions: - \(w = \sin(\sin(2x))\) - \(v = \sin(2x)\) - \(u = 2x\) Put it all together: \[ \frac{dy}{dx} = \cos(\sin(\sin(2x))) \cdot \cos(\sin(2x)) \cdot \cos(2x) \cdot 2 \] Or, simplifying: \[ \boxed{ \frac{dy}{dx} = 2 \cos(\sin(\sin(2x))) \cdot \cos(\sin(2x)) \cdot \cos(2x) } \]