1. **State the problem:** Differentiate the function $$f(t) = 6e^t \left(5\sqrt{t} + \frac{1}{2t^7}\right)$$ with respect to $$t$$. 2. **Recall the product rule:** If $$f(t) = u(t)v(t)$$, then $$f'(t) = u'(t)v(t) + u(t)v'(t)$$. 3. **Identify components:** Let $$u(t) = 6e^t$$ and $$v(t) = 5\sqrt{t} + \frac{1}{2t^7}$$. 4. **Differentiate $$u(t)$$:** $$u'(t) = 6e^t$$ because the derivative of $$e^t$$ is $$e^t$$. 5. **Rewrite $$v(t)$$ for easier differentiation:** $$v(t) = 5t^{1/2} + \frac{1}{2}t^{-7}$$. 6. **Differentiate $$v(t)$$:** $$v'(t) = 5 \cdot \frac{1}{2} t^{-1/2} + \frac{1}{2} \cdot (-7) t^{-8} = \frac{5}{2} t^{-1/2} - \frac{7}{2} t^{-8}$$. 7. **Apply the product rule:** $$f'(t) = u'(t)v(t) + u(t)v'(t) = 6e^t \left(5t^{1/2} + \frac{1}{2} t^{-7}\right) + 6e^t \left(\frac{5}{2} t^{-1/2} - \frac{7}{2} t^{-8}\right)$$. 8. **Factor out $$6e^t$$:** $$f'(t) = 6e^t \left(5t^{1/2} + \frac{1}{2} t^{-7} + \frac{5}{2} t^{-1/2} - \frac{7}{2} t^{-8}\right)$$. 9. **Final answer:** $$\boxed{f'(t) = 6e^t \left(5t^{1/2} + \frac{1}{2} t^{-7} + \frac{5}{2} t^{-1/2} - \frac{7}{2} t^{-8}\right)}$$