1. **Problem:** Given $x = t^2 - 1$ and $y = t^3 - 4$, find $\frac{d^2 y}{d x^2}$ at $t=1$. 2. **Step 1: Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$.** $$\frac{dy}{dt} = 3t^2, \quad \frac{dx}{dt} = 2t$$ 3. **Step 2: Find $\frac{dy}{dx}$ using the chain rule:** $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3t}{2}$$ 4. **Step 3: Differentiate $\frac{dy}{dx}$ with respect to $t$ to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:** $$\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$$ 5. **Step 4: Use the formula for the second derivative:** $$\frac{d^2 y}{d x^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t}$$ 6. **Step 5: Evaluate at $t=1$:** $$\frac{d^2 y}{d x^2}\bigg|_{t=1} = \frac{3}{4 \times 1} = \frac{3}{4}$$ **Final answer:** $\boxed{\frac{3}{4}}$ which corresponds to option (b).