1. **State the problem:** Find the second derivative of the function $$h(t) = (t^2 + 1) \sin t$$. 2. **Recall the product rule:** For two functions $u(t)$ and $v(t)$, the derivative is $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$. 3. **First derivative:** Let $u(t) = t^2 + 1$ and $v(t) = \sin t$. Calculate $u'(t) = 2t$. Calculate $v'(t) = \cos t$. Apply product rule: $$h'(t) = u'(t)v(t) + u(t)v'(t) = 2t \sin t + (t^2 + 1) \cos t$$. 4. **Second derivative:** Differentiate $h'(t)$ again: $$h''(t) = \frac{d}{dt}[2t \sin t] + \frac{d}{dt}[(t^2 + 1) \cos t]$$. Use product rule on each term: For $2t \sin t$: - $u = 2t$, $u' = 2$ - $v = \sin t$, $v' = \cos t$ $$\frac{d}{dt}[2t \sin t] = 2 \sin t + 2t \cos t$$. For $(t^2 + 1) \cos t$: - $u = t^2 + 1$, $u' = 2t$ - $v = \cos t$, $v' = -\sin t$ $$\frac{d}{dt}[(t^2 + 1) \cos t] = 2t \cos t + (t^2 + 1)(-\sin t) = 2t \cos t - (t^2 + 1) \sin t$$. 5. **Combine terms:** $$h''(t) = (2 \sin t + 2t \cos t) + (2t \cos t - (t^2 + 1) \sin t)$$ Simplify: $$h''(t) = 2 \sin t + 2t \cos t + 2t \cos t - (t^2 + 1) \sin t = 2 \sin t + 4t \cos t - (t^2 + 1) \sin t$$ Group like terms: $$h''(t) = (2 \sin t - (t^2 + 1) \sin t) + 4t \cos t = (2 - t^2 - 1) \sin t + 4t \cos t = (1 - t^2) \sin t + 4t \cos t$$. **Final answer:** $$\boxed{h''(t) = (1 - t^2) \sin t + 4t \cos t}$$