1. **Differentiate each given function:** 1. For $y = e^{x+2}$, use chain rule: $$\frac{dy}{dx} = e^{x+2} \times \frac{d}{dx}(x+2) = e^{x+2} \times 1 = e^{x+2}.$$ 2. For $y = e^{2x^2 - x}$, derivative: $$\frac{dy}{dx} = e^{2x^2 - x} \times \frac{d}{dx}(2x^2 - x) = e^{2x^2 - x} \times (4x - 1).$$ 3. For $y = e^{\sqrt{x+2}}$, set $u = \sqrt{x+2} = (x+2)^{1/2}$. Then $$\frac{dy}{dx} = e^u \times \frac{du}{dx} = e^{\sqrt{x+2}} \times \frac{1}{2\sqrt{x+2}}.$$ 4. For $y = \sqrt{e^{x^2}} + e^{\sqrt{x^2}} = e^{x^2/2} + e^{|x|}$ (since $\sqrt{x^2} = |x|$): - Derivative of $e^{x^2/2}$ is $e^{x^2/2} \times x$ by chain rule. - Derivative of $e^{|x|}$ is $e^{|x|} \times \frac{d}{dx} |x| = e^{|x|} \times \text{sgn}(x)$. Thus, $$\frac{dy}{dx} = xe^{x^2/2} + e^{|x|} \text{sgn}(x).$$ 5. For $y = e^{1/x^2} + \frac{1}{e^{x^2}} = e^{x^{-2}} + e^{-x^2}$: - Derivative of $e^{x^{-2}}$ is $e^{x^{-2}} \times (-2x^{-3}) = -2x^{-3}e^{1/x^2}$. - Derivative of $e^{-x^2}$ is $e^{-x^2} \times (-2x) = -2xe^{-x^2}$. Hence, $$\frac{dy}{dx} = -\frac{2}{x^3} e^{1/x^2} - 2x e^{-x^2}.$$ 6. For $y = e^{x^3 \ln x}$, set $u = x^3 \ln x$. Then $$\frac{dy}{dx} = e^u \times \frac{du}{dx}.$$ Compute $\frac{du}{dx} = 3x^2 \ln x + x^3 \times \frac{1}{x} = 3x^2 \ln x + x^2 = x^2(3\ln x + 1).$ So $$\frac{dy}{dx} = e^{x^3 \ln x} x^2 (3 \ln x + 1).$$ 7. For implicit differentiation of $e^{xy} + xy = 2$: Differentiate both sides w.r.t $x$: $$\frac{d}{dx}(e^{xy}) + \frac{d}{dx}(xy) = 0.$$ Using product and chain rules: $$e^{xy} (y + x \frac{dy}{dx}) + y + x \frac{dy}{dx} = 0.$$ Group terms with $\frac{dy}{dx}$: $$(e^{xy} x + x) \frac{dy}{dx} + e^{xy} y + y = 0.$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = - \frac{y (e^{xy} + 1)}{x(e^{xy} + 1)} = -\frac{y}{x}.$$ 2. **Calculate each integral:** 1. $$\int e^{3x+1} dx = \frac{1}{3} e^{3x+1} + C.$$ 2. $$\int x e^{x^2 -3} dx.$$ Let $u = x^2 - 3$, then $du = 2x dx \Rightarrow x dx = \frac{du}{2}$. So $$= \int e^u \frac{du}{2} = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2 - 3} + C.$$ 3. $$\int \frac{e^x}{e^x - 1} dx.$$ Substitute $u = e^x - 1$, $du = e^x dx$, so integral becomes $$\int \frac{1}{u} du = \ln|u| + C = \ln|e^x - 1| + C.$$ 4. $$\int \frac{e^{-1/x}}{x^2} dx.$$ Let $u = -1/x$, then $du = \frac{1}{x^2} dx$. Integral becomes $$\int e^u du = e^u + C = e^{-1/x} + C.$$ 5. $$\int (x + 3) e^{x^2 + 6x} dx.$$ Let $u = x^2 + 6x$, then $$du = (2x + 6) dx = 2(x + 3) dx \Rightarrow (x + 3) dx = \frac{du}{2}.$$ So integral is $$\int e^u \frac{du}{2} = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2 + 6x} + C.$$ 6. $$\int_0^1 e^{2x + 3} dx = e^3 \int_0^1 e^{2x} dx.$$ $$= e^3 \left[ \frac{1}{2} e^{2x} \right]_0^1 = e^3 \frac{1}{2} (e^2 - 1) = \frac{e^{5} - e^{3}}{2}.$$ 7. $$\int_1^2 \frac{e^{3/x}}{x^2} dx.$$ Let $u = 3/x$, so $du = -3/x^2 dx$ or $$-\frac{1}{3} du = \frac{1}{x^2} dx.$$ Integral becomes $$\int_{u=3}^{u=3/2} e^u \left(-\frac{1}{3}\right) du = -\frac{1}{3} \int_3^{3/2} e^u du = -\frac{1}{3} [e^u]_3^{3/2} = -\frac{1}{3} (e^{3/2} - e^{3}) = \frac{e^{3} - e^{3/2}}{3}.$$