1. **State the problem:** We need to find the derivative of the function $$f(x) = 10\sqrt{x^4 + 19}$$ at the point $$x = 3$$. 2. **Recall the formula:** The derivative of $$f(x) = 10(x^4 + 19)^{1/2}$$ can be found using the chain rule. If $$g(x) = (x^4 + 19)^{1/2}$$, then $$f(x) = 10g(x)$$ and $$f'(x) = 10g'(x)$$. 3. **Apply the chain rule:** $$g'(x) = \frac{1}{2}(x^4 + 19)^{-1/2} \cdot 4x^3 = \frac{2x^3}{\sqrt{x^4 + 19}}$$ 4. **Find $$f'(x)$$:** $$f'(x) = 10 \cdot \frac{2x^3}{\sqrt{x^4 + 19}} = \frac{20x^3}{\sqrt{x^4 + 19}}$$ 5. **Evaluate at $$x=3$$:** Calculate $$x^4 + 19 = 3^4 + 19 = 81 + 19 = 100$$. Then, $$f'(3) = \frac{20 \cdot 3^3}{\sqrt{100}} = \frac{20 \cdot 27}{10} = \frac{540}{10} = 54$$. **Final answer:** $$f'(3) = 54$$.