1. **State the problem:** We need to find the derivative of the function $$y = \sin^{-1}\left(\cos(2x^3 - 3x - 5)\right).$$ 2. **Recall the derivative formula:** If $$y = \sin^{-1}(u)$$ then $$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}.$$ Here, $$u = \cos(2x^3 - 3x - 5).$$ 3. **Find $$\frac{du}{dx}$$:** $$u = \cos(2x^3 - 3x - 5)$$ Using the chain rule, $$\frac{du}{dx} = -\sin(2x^3 - 3x - 5) \cdot \frac{d}{dx}(2x^3 - 3x - 5).$$ 4. **Compute $$\frac{d}{dx}(2x^3 - 3x - 5)$$:** $$\frac{d}{dx}(2x^3) = 6x^2,$$ $$\frac{d}{dx}(-3x) = -3,$$ $$\frac{d}{dx}(-5) = 0,$$ so, $$\frac{d}{dx}(2x^3 - 3x - 5) = 6x^2 - 3.$$ 5. **Therefore,** $$\frac{du}{dx} = -\sin(2x^3 - 3x - 5)(6x^2 - 3).$$ 6. **Substitute back into the derivative formula:** $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - \cos^2(2x^3 - 3x - 5)}} \cdot [-\sin(2x^3 - 3x - 5)(6x^2 - 3)].$$ 7. **Simplify the square root term:** Recall that $$\sin^2 \theta + \cos^2 \theta = 1,$$ so $$1 - \cos^2(\alpha) = \sin^2(\alpha).$$ Thus, $$\sqrt{1 - \cos^2(2x^3 - 3x - 5)} = |\sin(2x^3 - 3x - 5)|.$$ 8. **Final expression:** $$\frac{dy}{dx} = \frac{-\sin(2x^3 - 3x - 5)(6x^2 - 3)}{|\sin(2x^3 - 3x - 5)|} = -(6x^2 - 3) \frac{\sin(2x^3 - 3x - 5)}{|\sin(2x^3 - 3x - 5)|}.$$ 9. **Interpreting the fraction:** The fraction $$\frac{\sin(\theta)}{|\sin(\theta)|}$$ is the sign of $$\sin(\theta)$$, denoted $$\operatorname{sgn}(\sin(\theta))$$. So, $$\frac{dy}{dx} = -(6x^2 - 3) \operatorname{sgn}(\sin(2x^3 - 3x - 5)).$$ **Answer:** $$\boxed{\frac{dy}{dx} = -(6x^2 - 3) \operatorname{sgn}(\sin(2x^3 - 3x - 5))}.$$