1. **Problem statement:** Find the derivative of the function $$f(x) = \tan^4 \left( \sin^2 \left( x^3 + 2x \right) \right).$$ 2. **Formula and rules:** We will use the chain rule multiple times. The chain rule states that if $$y = f(g(x)),$$ then $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x).$$ 3. **Step-by-step differentiation:** - Let $$u = \sin^2 \left( x^3 + 2x \right),$$ so $$f(x) = \tan^4(u).$$ - Derivative of $$\tan^4(u)$$ with respect to $$u$$ is $$4 \tan^3(u) \cdot \sec^2(u).$$ - Next, differentiate $$u = \sin^2(v)$$ where $$v = x^3 + 2x.$$ Using chain rule again: $$\frac{du}{dv} = 2 \sin(v) \cos(v) = \sin(2v)$$ (using the double angle identity). - Finally, differentiate $$v = x^3 + 2x$$: $$\frac{dv}{dx} = 3x^2 + 2.$$ 4. **Combine all parts:** $$\frac{df}{dx} = 4 \tan^3(u) \sec^2(u) \cdot \sin(2v) \cdot (3x^2 + 2).$$ Substitute back $$u$$ and $$v$$: $$\frac{df}{dx} = 4 \tan^3 \left( \sin^2 \left( x^3 + 2x \right) \right) \sec^2 \left( \sin^2 \left( x^3 + 2x \right) \right) \sin \left( 2 \left( x^3 + 2x \right) \right) (3x^2 + 2).$$ 5. **Simplify the coefficient:** $$4 (3x^2 + 2) = 12x^2 + 8.$$ 6. **Final answer:** $$\boxed{(12x^2 + 8) \tan^3 \left( \sin^2 \left( x^3 + 2x \right) \right) \sec^2 \left( \sin^2 \left( x^3 + 2x \right) \right) \sin \left( 2x^3 + 4x \right)}.$$