Let \( f(x)=-4 \cos \left(\sin \left(x^{3}\right)\right) \) Then \( f^{\prime}(x)= \)

To find the derivative \( f'(x) \) of the function \( f(x) = -4 \cos(\sin(x^3)) \), we will apply the chain rule. First, we differentiate the outer function, which is \( -4 \cos(u) \) where \( u = \sin(x^3) \). The derivative of \( -4 \cos(u) \) is \( 4 \sin(u) \frac{du}{dx} \). Now, we differentiate \( u = \sin(x^3) \). By applying the chain rule again, the derivative of \( \sin(v) \) where \( v = x^3 \) is \( \cos(v) \cdot \frac{dv}{dx} \). Differentiating \( v = x^3 \) gives us \( 3x^2 \). Putting it all together, we have: 1. Differentiate the outer part: \( -4 \cos(\sin(x^3)) \) gives \( 4 \sin(\sin(x^3)) \). 2. Differentiate the middle part: \( \sin(x^3) \) gives \( \cos(x^3) \cdot 3x^2 \). Putting these derivatives together, we find: \[ f'(x) = 4 \sin(\sin(x^3)) \cdot \cos(x^3) \cdot 3x^2. \] So, the final answer is: \[ f'(x) = 12x^2 \sin(\sin(x^3)) \cos(x^3). \]