1. We are given the function $$f(t) = 5t^2 - \frac{1}{t^3} + 3t - \sqrt{t} + 1$$ and asked to find the second derivative $$f''(t)$$ at $$t=1$$. 2. First, recall the rules for derivatives: - Power rule: $$\frac{d}{dt} t^n = n t^{n-1}$$ - Derivative of $$t^{-n}$$ is $$-n t^{-n-1}$$ - Derivative of $$\sqrt{t} = t^{1/2}$$ is $$\frac{1}{2} t^{-1/2}$$ 3. Find the first derivative $$f'(t)$$: $$f'(t) = \frac{d}{dt} \left(5t^2\right) - \frac{d}{dt} \left(t^{-3}\right) + \frac{d}{dt} \left(3t\right) - \frac{d}{dt} \left(t^{1/2}\right) + \frac{d}{dt} (1)$$ Calculate each term: - $$\frac{d}{dt} 5t^2 = 10t$$ - $$\frac{d}{dt} t^{-3} = -3 t^{-4}$$ - $$\frac{d}{dt} 3t = 3$$ - $$\frac{d}{dt} t^{1/2} = \frac{1}{2} t^{-1/2}$$ - $$\frac{d}{dt} 1 = 0$$ So, $$f'(t) = 10t + 3 t^{-4} + 3 - \frac{1}{2} t^{-1/2}$$ 4. Now find the second derivative $$f''(t)$$ by differentiating $$f'(t)$$: $$f''(t) = \frac{d}{dt} (10t) + \frac{d}{dt} (3 t^{-4}) + \frac{d}{dt} (3) - \frac{d}{dt} \left(\frac{1}{2} t^{-1/2}\right)$$ Calculate each term: - $$\frac{d}{dt} 10t = 10$$ - $$\frac{d}{dt} 3 t^{-4} = 3 \times (-4) t^{-5} = -12 t^{-5}$$ - $$\frac{d}{dt} 3 = 0$$ - $$\frac{d}{dt} \left(\frac{1}{2} t^{-1/2}\right) = \frac{1}{2} \times (-\frac{1}{2}) t^{-3/2} = -\frac{1}{4} t^{-3/2}$$ So, $$f''(t) = 10 - 12 t^{-5} + \frac{1}{4} t^{-3/2}$$ 5. Evaluate $$f''(t)$$ at $$t=1$$: $$f''(1) = 10 - 12 (1)^{-5} + \frac{1}{4} (1)^{-3/2} = 10 - 12 + \frac{1}{4} = -2 + \frac{1}{4} = -\frac{7}{4} = -1.75$$ Final answer: $$\boxed{f''(1) = -\frac{7}{4}}$$