1. **Problem statement:** Find the derivative $ \frac{d}{dx}(\ln \ln 2x) $ and then find $ \frac{dy}{dx} $ for $ y = \ln \ln \ln 2x $. 2. **Recall the chain rule:** If $ y = f(g(x)) $, then $ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) $. 3. **Step 1: Differentiate $ \ln \ln 2x $:** Let $ u = \ln 2x $, then $ \ln \ln 2x = \ln u $. Using the chain rule: $$ \frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx} $$ 4. **Find $ \frac{du}{dx} $:** Since $ u = \ln 2x $, $$ \frac{du}{dx} = \frac{1}{2x} \cdot \frac{d}{dx}(2x) = \frac{1}{2x} \cdot 2 = \frac{2}{2x} = \frac{1}{x} $$ 5. **Substitute back:** $$ \frac{d}{dx}(\ln \ln 2x) = \frac{1}{\ln 2x} \cdot \frac{1}{x} = \frac{1}{x \ln 2x} $$ 6. **Step 2: Differentiate $ y = \ln \ln \ln 2x $:** Let $ v = \ln 2x $, then $ w = \ln v $, so $ y = \ln w $. Using the chain rule: $$ \frac{dy}{dx} = \frac{1}{w} \cdot \frac{dw}{dx} $$ 7. **Find $ \frac{dw}{dx} $:** Since $ w = \ln v $, $$ \frac{dw}{dx} = \frac{1}{v} \cdot \frac{dv}{dx} $$ 8. **Find $ \frac{dv}{dx} $:** Since $ v = \ln 2x $, $$ \frac{dv}{dx} = \frac{1}{2x} \cdot 2 = \frac{1}{x} $$ 9. **Substitute back:** $$ \frac{dw}{dx} = \frac{1}{\ln 2x} \cdot \frac{1}{x} = \frac{1}{x \ln 2x} $$ 10. **Final expression for $ \frac{dy}{dx} $:** $$ \frac{dy}{dx} = \frac{1}{\ln \ln 2x} \cdot \frac{1}{x \ln 2x} = \frac{1}{x (\ln 2x) (\ln \ln 2x)} $$ 11. **Summary:** - $ \frac{d}{dx}(\ln \ln 2x) = \frac{1}{x \ln 2x} $ - $ \frac{dy}{dx} = \frac{1}{x (\ln 2x) (\ln \ln 2x)} $ for $ y = \ln \ln \ln 2x $.