1. **State the problem:** We need to find the derivative of the function $$f(x) = \sin^{-1}(8x^3)$$. 2. **Recall the formula:** The derivative of $$\sin^{-1}(u)$$ with respect to $$x$$ is $$\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$. 3. **Identify the inner function:** Here, $$u = 8x^3$$. 4. **Compute the derivative of the inner function:** $$\frac{du}{dx} = 24x^2$$. 5. **Apply the chain rule:** $$ f'(x) = \frac{1}{\sqrt{1-(8x^3)^2}} \cdot 24x^2 = \frac{24x^2}{\sqrt{1-64x^6}}$$ 6. **Final answer:** $$ f'(x) = \frac{24x^2}{\sqrt{1-64x^6}}$$ This derivative tells us the rate of change of the inverse sine of $$8x^3$$ with respect to $$x$$.