1. **Problem Statement:** Find the first, second, and third derivatives of the function $$y(x) = e^{6x^2 + 3x - 2}$$ with respect to $$x$$. 2. **Formula and Rules:** The derivative of an exponential function $$e^{u(x)}$$ is given by $$\frac{dy}{dx} = e^{u(x)} \cdot u'(x)$$ where $$u'(x)$$ is the derivative of the exponent function. 3. **Step 1: Define $$u(x)$$:** $$u(x) = 6x^2 + 3x - 2$$ 4. **Step 2: Compute $$u'(x)$$:** $$u'(x) = \frac{d}{dx}(6x^2 + 3x - 2) = 12x + 3$$ 5. **Step 3: First derivative $$y'$$:** $$y' = e^{6x^2 + 3x - 2} \cdot (12x + 3)$$ 6. **Step 4: Second derivative $$y''$$:** Use product rule: $$y'' = \frac{d}{dx}[e^{u(x)}(12x + 3)] = e^{u(x)}(12x + 3)^2 + e^{u(x)} \cdot 12$$ Simplify: $$y'' = e^{6x^2 + 3x - 2} \left[(12x + 3)^2 + 12\right]$$ 7. **Step 5: Third derivative $$y'''$$:** Again use product rule: $$y''' = \frac{d}{dx}\left[e^{u(x)} \left((12x + 3)^2 + 12\right)\right]$$ $$= e^{u(x)} (12x + 3) \left((12x + 3)^2 + 12\right) + e^{u(x)} \cdot 24 (12x + 3)$$ Simplify: $$y''' = e^{6x^2 + 3x - 2} (12x + 3) \left[(12x + 3)^2 + 36\right]$$ **Final answers:** $$y' = e^{6x^2 + 3x - 2} (12x + 3)$$ $$y'' = e^{6x^2 + 3x - 2} \left[(12x + 3)^2 + 12\right]$$ $$y''' = e^{6x^2 + 3x - 2} (12x + 3) \left[(12x + 3)^2 + 36\right]$$ These derivatives show how the function changes at different rates and orders.