1. **State the problem:** Find the derivative $y'$ if $y = -e^{\csc(x^2)}$. 2. **Recall the chain rule:** If $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$. 3. **Identify the outer and inner functions:** - Outer function: $f(u) = -e^u$ where $u = \csc(x^2)$. - Inner function: $u = \csc(x^2)$. 4. **Differentiate the outer function:** $$f'(u) = -e^u$$ 5. **Differentiate the inner function $u = \csc(x^2)$:** Recall that $\frac{d}{dx} \csc(v) = -\csc(v) \cot(v) \cdot v'$. Here, $v = x^2$, so $v' = 2x$. Therefore, $$u' = -\csc(x^2) \cot(x^2) \cdot 2x$$ 6. **Apply the chain rule:** $$y' = f'(u) \cdot u' = -e^{\csc(x^2)} \cdot \left(-\csc(x^2) \cot(x^2) \cdot 2x\right)$$ 7. **Simplify:** $$y' = 2x e^{\csc(x^2)} \csc(x^2) \cot(x^2)$$ **Final answer:** $$y' = 2x e^{\csc(x^2)} \csc(x^2) \cot(x^2)$$