1. Problem statement: Find the derivative of $y = \sin^{-1}(\cos(2x^3 - 3x - 5))$. 2. Strategy: We treat this as a composition $y = f(g(h(x)))$ where $h(x)=2x^3-3x-5$, $g(t)=\cos t$, and $f(u)=\sin^{-1}u$. 3. Compute inner derivatives: $h'(x)=6x^2-3$. 4. Compute derivative of $g$: $g'(t)=-\sin t$, so $g'(h(x))=-\sin(2x^3-3x-5)$. 5. Compute derivative of $f$: $f'(u)=\dfrac{1}{\sqrt{1-u^2}}$. 6. Apply the chain rule: $y'=f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)$. 7. Substitute the expressions: $y'=\dfrac{1}{\sqrt{1-\cos^2(2x^3-3x-5)}}\cdot\big(-\sin(2x^3-3x-5)\big)\cdot(6x^2-3)$. 8. Simplify the product: $y'=-\dfrac{(6x^2-3)\sin(2x^3-3x-5)}{\sqrt{1-\cos^2(2x^3-3x-5)}}$. 9. Optional simplification: since $\sqrt{1-\cos^2 t}=|\sin t|$, for points where $\sin(2x^3-3x-5)\neq 0$ we can write $y'=-(6x^2-3)\,\operatorname{sgn}(\sin(2x^3-3x-5))$. 10. Final answer: $y'=-\dfrac{(6x^2-3)\sin(2x^3-3x-5)}{\sqrt{1-\cos^2(2x^3-3x-5)}}$.