Here are the derivatives for the remaining functions: 2. $f(x) = \frac{\ln(x)}{x^2 \sin(x)}$ Step 1: Identify $u$ and $v$ for the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Let $u = \ln(x)$ and $v = x^2 \sin(x)$. Step 2: Find the derivatives of $u$ and $v$. For $u$: $u' = \frac{1}{x}$. For $v$, use the product rule: $v' = \frac{d}{dx}(x^2)\sin(x) + x^2\frac{d}{dx}(\sin(x)) = 2x\sin(x) + x^2\cos(x)$. Step 3: Apply the quotient rule. $$ f'(x) = \frac{\left(\frac{1}{x}\right)(x^2 \sin(x)) - (\ln(x))(2x\sin(x) + x^2\cos(x))}{(x^2 \sin(x))^2} $$ Step 4: Simplify the expression. $$ f'(x) = \frac{x\sin(x) - 2x\ln(x)\sin(x) - x^2\ln(x)\cos(x)}{x^4 \sin^2(x)} $$ Factor out $x$ from the numerator. $$ f'(x) = \frac{x(\sin(x) - 2\ln(x)\sin(x) - x\ln(x)\cos(x))}{x^4 \sin^2(x)} $$ Cancel one $x$ from numerator and denominator. $$ f'(x) = \boxed{\frac{\sin(x) - 2\ln(x)\sin(x) - x\ln(x)\cos(x)}{x^3 \sin^2(x)}} $$ --- 4. $y = \frac{\tan(x)}{e^{3x}}$ Step 1: Identify $u$ and $v$ for the quotient rule. Let $u = \tan(x)$ and $v = e^{3x}$. Step 2: Find the derivatives of $u$ and $v$. For $u$: $u' = \sec^2(x)$. For $v$, use the chain rule: $v' = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x}$. Step 3: Apply the quotient rule. $$ \frac{dy}{dx} = \frac{(\sec^2(x))(e^{3x}) - (\tan(x))(3e^{3x})}{(e^{3x})^2} $$ Step 4: Simplify the expression. $$ \frac{dy}{dx} = \frac{e^{3x}(\sec^2(x) - 3\tan(x))}{e^{6x}} $$ Cancel $e^{3x}$ from numerator and denominator. $$ \frac{dy}{dx} = \boxed{\frac{\sec^2(x) - 3\tan(x)}{e^{3x}}} $$ --- 6. $f(x) = \frac{2x}{x^2}$ Step 1: Simplify the function first. $$ f(x) = \frac{2}{x} = 2x^{-1} $$ Step 2: Differentiate the simplified function using the power rule. $$ f'(x) = 2 \cdot (-1)x^{-1-1} $$ $$ f'(x) = -2x^{-2} $$ $$ f'(x) = \boxed{-\frac{2}{x^2}} $$ --- 8. $y = \frac{\ln(x^2 + 1)}{\sqrt{x}}$ Step 1: Identify $u$ and $v$ for the quotient rule. Let $u = \ln(x^2 + 1)$ and $v = \sqrt{x} = x^{1/2}$. Step 2: Find the derivatives of $u$ and $v$. For $u$, use the chain rule: $u' = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{x^2 + 1}$. For $v$: $v' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. Step 3: Apply the quotient rule. $$ \frac{dy}{dx} = \frac{\left(\frac{2x}{x^2 + 1}\right)(\sqrt{x}) - (\ln(x^2 + 1))\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2} $$ Step 4: Simplify the expression. Multiply the numerator and denominator by $2\sqrt{x}(x^2+1)$ to clear fractions in the numerator. $$ \frac{dy}{dx} = \frac{\frac{2x\sqrt{x}}{x^2 + 1} - \frac{\ln(x^2 + 1)}{2\sqrt{x}}}{x} \cdot \frac{2\sqrt{x}(x^2+1)}{2\sqrt{x}(x^2+1)} $$ $$ \frac{dy}{dx} = \frac{2x\sqrt{x}(2\sqrt{x}) - \ln(x^2 + 1)(x^2+1)}{x \cdot 2\sqrt{x}(x^2+1)} $$ $$ \frac{dy}{dx} = \frac{4x^2 - (x^2+1)\ln(x^2 + 1)}{2x^{3/2}(x^2+1)} $$ $$ \frac{dy}{dx} = \boxed{\frac{4x^2 - (x^2+1)\ln(x^2 + 1)}{2x\sqrt{x}(x^2+1)}} $$ --- 10. $f(x) = \frac{e^x \ln x}{x^2 + 1}$ Step 1: Identify $u$ and $v$ for the quotient rule. Let $u = e^x \ln x$ and $v = x^2 + 1$. Step 2: Find the derivatives of $u$ and $v$. For $u$, use the product rule: $u' = \frac{d}{dx}(e^x)\ln x + e^x\frac{d}{dx}(\ln x) = e^x \ln x + e^x \cdot \frac{1}{x} = e^x\left(\ln x + \frac{1}{x}\right)$. For $v$: $v' = 2x$. Step 3: Apply the quotient rule. $$ f'(x) = \frac{\left(e^x\left(\ln x + \frac{1}{x}\right)\right)(x^2 + 1) - (e^x \ln x)(2x)}{(x^2 + 1)^2} $$ Step 4: Simplify the expression. Factor out $e^x$ from the numerator. $$ f'(x) = \frac{e^x\left[\left(\ln x + \frac{1}{x}\right)(x^2 + 1) - 2x \ln x\right]}{(x^2 + 1)^2} $$ Expand the term in the brackets. $$ f'(x) = \frac{e^x\left[x^2 \ln x + \ln x + x + \frac{1}{x} - 2x \ln x\right]}{(x^2 + 1)^2} $$ Combine like terms. $$ f'(x) = \boxed{\frac{e^x\left[x^2 \ln x - 2x \ln x + \ln x + x + \frac{1}{x}\right]}{(x^2 + 1)^2}} $$ That's 2 down. 3 left today — send the next one.