1. We are asked to find the derivative of the function \( y = \frac{\sqrt{9+3x-x^{2}}}{e^{x}} \). 2. Recall the quotient rule for derivatives: if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^{2}} \). 3. Let \( u = \sqrt{9+3x-x^{2}} = (9+3x-x^{2})^{\frac{1}{2}} \) and \( v = e^{x} \). 4. Compute \( u' \): Using the chain rule, $$ u' = \frac{1}{2}(9+3x-x^{2})^{-\frac{1}{2}} \cdot (3 - 2x) = \frac{3 - 2x}{2\sqrt{9+3x-x^{2}}} $$ 5. Compute \( v' = e^{x} \). 6. Apply the quotient rule: $$ y' = \frac{\frac{3 - 2x}{2\sqrt{9+3x-x^{2}}} \cdot e^{x} - \sqrt{9+3x-x^{2}} \cdot e^{x}}{(e^{x})^{2}} $$ 7. Simplify numerator: $$ e^{x} \left( \frac{3 - 2x}{2\sqrt{9+3x-x^{2}}} - \sqrt{9+3x-x^{2}} \right) $$ 8. Divide by \( e^{2x} \) to get: $$ y' = e^{-x} \left( \frac{3 - 2x}{2\sqrt{9+3x-x^{2}}} - \sqrt{9+3x-x^{2}} \right) $$ 9. Combine terms inside parentheses over common denominator: $$ \frac{3 - 2x - 2(9+3x-x^{2})}{2\sqrt{9+3x-x^{2}}} = \frac{3 - 2x - 18 - 6x + 2x^{2}}{2\sqrt{9+3x-x^{2}}} = \frac{2x^{2} - 8x - 15}{2\sqrt{9+3x-x^{2}}} $$ 10. Final derivative: $$ y' = e^{-x} \cdot \frac{2x^{2} - 8x - 15}{2\sqrt{9+3x-x^{2}}} $$ --- Since the user asked for the derivative of the first function only (per guest rule), we stop here. "slug": "derivative sqrt", "subject": "calculus", "desmos": {"latex": "y=\frac{\sqrt{9+3x-x^{2}}}{e^{x}}","features": {"intercepts": true,"extrema": true}}, "q_count": 2