1. **Problem Statement:** Find the first, second, and third derivatives of the functions: i. $y(x) = e^{6x^2 + 3x - 2}$ ii. $y(x) = e^{3x - 2}$ iii. $y(x) = e^{kx}$ 2. **Formula and Rules:** The derivative of $e^{u(x)}$ with respect to $x$ is given by the chain rule: $$\frac{dy}{dx} = e^{u(x)} \cdot u'(x)$$ For higher order derivatives, apply the product and chain rules repeatedly. 3. **Solution for i. $y = e^{6x^2 + 3x - 2}$:** - Let $u = 6x^2 + 3x - 2$ - Then $u' = 12x + 3$ **First derivative:** $$y' = e^{u} \cdot u' = e^{6x^2 + 3x - 2} (12x + 3)$$ **Second derivative:** Apply product rule: $$y'' = \frac{d}{dx}[e^{u}(12x + 3)] = e^{u} u' (12x + 3) + e^{u} (12) = e^{6x^2 + 3x - 2} [(12x + 3)^2 + 12]$$ **Third derivative:** Again, differentiate $y''$: $$y''' = \frac{d}{dx} \left(e^{u} [(12x + 3)^2 + 12] \right)$$ Using product rule: $$y''' = e^{u} u' [(12x + 3)^2 + 12] + e^{u} \cdot 2(12x + 3)(12)$$ Simplify: $$y''' = e^{6x^2 + 3x - 2} \left[(12x + 3)^3 + 12(12x + 3) + 24(12x + 3) \right] = e^{6x^2 + 3x - 2} \left[(12x + 3)^3 + 36(12x + 3) \right]$$ 4. **Solution for ii. $y = e^{3x - 2}$:** - Let $u = 3x - 2$ - Then $u' = 3$ **First derivative:** $$y' = e^{3x - 2} \cdot 3 = 3 e^{3x - 2}$$ **Second derivative:** $$y'' = \frac{d}{dx} (3 e^{3x - 2}) = 3 \cdot 3 e^{3x - 2} = 9 e^{3x - 2}$$ **Third derivative:** $$y''' = \frac{d}{dx} (9 e^{3x - 2}) = 9 \cdot 3 e^{3x - 2} = 27 e^{3x - 2}$$ 5. **Solution for iii. $y = e^{kx}$:** - Let $u = kx$ - Then $u' = k$ **First derivative:** $$y' = e^{kx} \cdot k = k e^{kx}$$ **Second derivative:** $$y'' = \frac{d}{dx} (k e^{kx}) = k^2 e^{kx}$$ **Third derivative:** $$y''' = \frac{d}{dx} (k^2 e^{kx}) = k^3 e^{kx}$$ **Summary:** - For $y = e^{6x^2 + 3x - 2}$, derivatives involve chain and product rules with polynomial factors. - For $y = e^{3x - 2}$, derivatives scale by powers of 3. - For $y = e^{kx}$, the $n$th derivative is $k^n e^{kx}$. This matches the program logic where the $n$th derivative of $e^{ax}$ is $a^n e^{ax}$.