1. **Problem:** Find the equation of the tangent line to the function $$f(x) = x^2 \sqrt{16 - x^2}$$ at $$x = 1$$. 2. **Step 1: Find the derivative $$f'(x)$$.** Given $$f(x) = x^2 (16 - x^2)^{1/2}$$, use the product rule: $$f'(x) = \frac{d}{dx}[x^2] \cdot (16 - x^2)^{1/2} + x^2 \cdot \frac{d}{dx}[(16 - x^2)^{1/2}]$$ 3. **Calculate each derivative:** $$\frac{d}{dx}[x^2] = 2x$$ $$\frac{d}{dx}[(16 - x^2)^{1/2}] = \frac{1}{2}(16 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{(16 - x^2)^{1/2}}$$ 4. **Substitute back:** $$f'(x) = 2x (16 - x^2)^{1/2} + x^2 \left(-\frac{x}{(16 - x^2)^{1/2}}\right) = 2x (16 - x^2)^{1/2} - \frac{x^3}{(16 - x^2)^{1/2}}$$ 5. **Combine terms over common denominator:** $$f'(x) = \frac{2x (16 - x^2) - x^3}{(16 - x^2)^{1/2}}$$ 6. **Simplify numerator:** $$2x (16 - x^2) - x^3 = 32x - 2x^3 - x^3 = 32x - 3x^3$$ So, $$f'(x) = \frac{32x - 3x^3}{(16 - x^2)^{1/2}}$$ 7. **Evaluate $$f'(1)$$:** $$f'(1) = \frac{32(1) - 3(1)^3}{\sqrt{16 - 1}} = \frac{32 - 3}{\sqrt{15}} = \frac{29}{\sqrt{15}}$$ 8. **Find $$f(1)$$:** $$f(1) = 1^2 \cdot \sqrt{16 - 1} = \sqrt{15}$$ 9. **Equation of tangent line:** Using point-slope form: $$y - f(1) = f'(1)(x - 1)$$ $$y - \sqrt{15} = \frac{29}{\sqrt{15}} (x - 1)$$ 10. **Final answer:** $$\boxed{y = \frac{29}{\sqrt{15}} (x - 1) + \sqrt{15}}$$