1. Find the derivative of each function: 1) Given $y = \ln(\sin^{-1}(2x))$. Step 1: Use the chain rule. Let $u = \sin^{-1}(2x)$, then $y = \ln(u)$. Step 2: Derivative of $\ln(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$. Step 3: Derivative of $u = \sin^{-1}(2x)$ is $\frac{2}{\sqrt{1-(2x)^2}} = \frac{2}{\sqrt{1-4x^2}}$. Step 4: Combine: $$\frac{dy}{dx} = \frac{1}{\sin^{-1}(2x)} \cdot \frac{2}{\sqrt{1-4x^2}}$$ 2) Given $y = 7^{\ln(7x)}$. Step 1: Rewrite using exponentials: $y = e^{\ln(7x) \ln 7}$. Step 2: Differentiate: $$\frac{dy}{dx} = y \cdot \frac{d}{dx}(\ln(7x) \ln 7) = 7^{\ln(7x)} \cdot \ln 7 \cdot \frac{1}{7x} \cdot 7 = 7^{\ln(7x)} \cdot \frac{\ln 7}{x}$$ 3) Given $y = \sec^{-1}(\ln x - e^{4x})$. Step 1: Let $u = \ln x - e^{4x}$. Step 2: Derivative of $\sec^{-1}(u)$ is $\frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}$. Step 3: Compute $\frac{du}{dx} = \frac{1}{x} - 4 e^{4x}$. Step 4: Combine: $$\frac{dy}{dx} = \frac{1}{|\ln x - e^{4x}| \sqrt{(\ln x - e^{4x})^2 - 1}} \left( \frac{1}{x} - 4 e^{4x} \right)$$ 4) Given $y = \log_3(\sec 3x)$. Step 1: Change base formula: $y = \frac{\ln(\sec 3x)}{\ln 3}$. Step 2: Differentiate: $$\frac{dy}{dx} = \frac{1}{\ln 3} \cdot \frac{1}{\sec 3x} \cdot \sec 3x \tan 3x \cdot 3 = \frac{3 \tan 3x}{\ln 3}$$ II. Evaluate the integrals using substitution: 1) $\int \frac{\sec(2x) \tan(2x)}{4 + \sec(2x)} dx$ Step 1: Let $u = 4 + \sec(2x)$, then $du = 2 \sec(2x) \tan(2x) dx$. Step 2: Rewrite integral: $$\int \frac{\sec(2x) \tan(2x)}{4 + \sec(2x)} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|4 + \sec(2x)| + C$$ 2) $\int \frac{2^{\tan^{-1} x}}{1 + x^2} dx$ Step 1: Let $u = \tan^{-1} x$, then $du = \frac{1}{1 + x^2} dx$. Step 2: Integral becomes: $$\int 2^u du = \frac{2^u}{\ln 2} + C = \frac{2^{\tan^{-1} x}}{\ln 2} + C$$ 3) $\int e^{\csc(4x)} \csc(4x) \cot(4x) dx$ Step 1: Let $u = \csc(4x)$, then $du = -4 \csc(4x) \cot(4x) dx$. Step 2: Rewrite integral: $$\int e^u \csc(4x) \cot(4x) dx = -\frac{1}{4} \int e^u du = -\frac{1}{4} e^u + C = -\frac{1}{4} e^{\csc(4x)} + C$$ 4) $\int \frac{\sin(3x)}{4 + \cos^2(3x)} dx$ Step 1: Let $u = \cos(3x)$, then $du = -3 \sin(3x) dx$. Step 2: Rewrite integral: $$\int \frac{\sin(3x)}{4 + \cos^2(3x)} dx = -\frac{1}{3} \int \frac{du}{4 + u^2} = -\frac{1}{3} \cdot \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C = -\frac{1}{6} \tan^{-1} \left( \frac{\cos(3x)}{2} \right) + C$$ 5) $\int \frac{dx}{(x+5) \sqrt{x^2 + 10x}}$ Step 1: Complete the square inside the root: $x^2 + 10x = (x+5)^2 - 25$. Step 2: Let $u = x + 5$, then $du = dx$. Step 3: Integral becomes: $$\int \frac{du}{u \sqrt{u^2 - 25}}$$ Step 4: Use substitution $u = 5 \sec \theta$, then $du = 5 \sec \theta \tan \theta d\theta$. Step 5: Substitute: $$\int \frac{5 \sec \theta \tan \theta d\theta}{5 \sec \theta \sqrt{25 \sec^2 \theta - 25}} = \int \frac{5 \sec \theta \tan \theta d\theta}{5 \sec \theta \cdot 5 \tan \theta} = \int \frac{5 \sec \theta \tan \theta d\theta}{25 \sec \theta \tan \theta} = \int \frac{d\theta}{5} = \frac{\theta}{5} + C$$ Step 6: Back-substitute $\theta = \sec^{-1} \frac{u}{5} = \sec^{-1} \frac{x+5}{5}$. Final answer: $$\frac{1}{5} \sec^{-1} \left( \frac{x+5}{5} \right) + C$$ 6) $\int \frac{e^x}{\sqrt{7 - e^{2x}}} dx$ Step 1: Let $u = e^x$, then $du = e^x dx$. Step 2: Rewrite integral: $$\int \frac{du}{\sqrt{7 - u^2}} = \sin^{-1} \left( \frac{u}{\sqrt{7}} \right) + C = \sin^{-1} \left( \frac{e^x}{\sqrt{7}} \right) + C$$ III. Use integration by parts: 1) $\int x^3 \sin(3x) dx$ Step 1: Let $u = x^3$, $dv = \sin(3x) dx$. Step 2: Then $du = 3x^2 dx$, $v = -\frac{1}{3} \cos(3x)$. Step 3: Apply integration by parts: $$\int x^3 \sin(3x) dx = -\frac{1}{3} x^3 \cos(3x) + \int x^2 \cos(3x) dx$$ Step 4: Repeat integration by parts on $\int x^2 \cos(3x) dx$ similarly until fully integrated. 2) $\int \sin^{-1}(5x) dx$ Step 1: Let $u = \sin^{-1}(5x)$, $dv = dx$. Step 2: Then $du = \frac{5}{\sqrt{1 - 25x^2}} dx$, $v = x$. Step 3: Integration by parts: $$x \sin^{-1}(5x) - \int \frac{5x}{\sqrt{1 - 25x^2}} dx$$ Step 4: Use substitution $w = 1 - 25x^2$ to solve the remaining integral. 3) $\int (x^8 + 8x) \ln(8x) dx$ Step 1: Let $u = \ln(8x)$, $dv = (x^8 + 8x) dx$. Step 2: Then $du = \frac{1}{x} dx$, $v = \frac{x^9}{9} + 4x^2$. Step 3: Integration by parts: $$\left( \frac{x^9}{9} + 4x^2 \right) \ln(8x) - \int \left( \frac{x^9}{9} + 4x^2 \right) \frac{1}{x} dx = \left( \frac{x^9}{9} + 4x^2 \right) \ln(8x) - \int \left( \frac{x^8}{9} + 4x \right) dx$$ Step 4: Integrate the remaining polynomial: $$- \left( \frac{x^9}{81} + 2x^2 \right) + C$$ Final answer: $$\left( \frac{x^9}{9} + 4x^2 \right) \ln(8x) - \frac{x^9}{81} - 2x^2 + C$$