1. **Problem:** Find the derivative of $f(x) = (3x^3 + 3x - 1)^{10}$. Step 1: Use the chain rule. Let $u = 3x^3 + 3x - 1$, then $f(x) = u^{10}$. Step 2: Derivative of $u^{10}$ is $10u^9 \cdot u'$. Step 3: Compute $u' = \frac{d}{dx}(3x^3 + 3x - 1) = 9x^2 + 3$. Step 4: Combine results: $$f'(x) = 10(3x^3 + 3x - 1)^9 (9x^2 + 3)$$ --- 2. **Problem:** Find the derivative of $f(x) = 3e^{3x} - \ln x^2$. Step 1: Derivative of $3e^{3x}$ is $3 \cdot 3e^{3x} = 9e^{3x}$. Step 2: Simplify $\ln x^2 = 2 \ln x$. Step 3: Derivative of $2 \ln x$ is $\frac{2}{x}$. Step 4: Combine results: $$f'(x) = 9e^{3x} - \frac{2}{x}$$ --- 3. **Problem:** Find the derivative of $f(x) = (x + 2)^2 (x^2 + 3)$. Step 1: Use product rule: $f'(x) = g'(x)h(x) + g(x)h'(x)$ where $g(x) = (x+2)^2$, $h(x) = x^2 + 3$. Step 2: Compute $g'(x) = 2(x+2)$. Step 3: Compute $h'(x) = 2x$. Step 4: Substitute: $$f'(x) = 2(x+2)(x^2 + 3) + (x+2)^2 (2x)$$ Step 5: Factor if desired or leave as is. --- 4. **Problem:** Find the derivative of $f(x) = \sin^3(2x) - 3 \cos(5x^2)$. Step 1: Rewrite $\sin^3(2x) = (\sin(2x))^3$. Step 2: Use chain rule: $$\frac{d}{dx} (\sin(2x))^3 = 3(\sin(2x))^2 \cdot \cos(2x) \cdot 2 = 6 \sin^2(2x) \cos(2x)$$ Step 3: Derivative of $-3 \cos(5x^2)$ is: $$-3 \cdot (-\sin(5x^2)) \cdot 10x = 30x \sin(5x^2)$$ Step 4: Combine results: $$f'(x) = 6 \sin^2(2x) \cos(2x) + 30x \sin(5x^2)$$ --- 5. **Problem:** Find the derivative of $f(x) = \tan^{1/x}(x)$. Step 1: Rewrite as $f(x) = (\tan x)^{1/x}$. Step 2: Take natural log: $$\ln f(x) = \frac{1}{x} \ln(\tan x)$$ Step 3: Differentiate both sides: $$\frac{f'(x)}{f(x)} = \frac{d}{dx} \left( \frac{1}{x} \ln(\tan x) \right)$$ Step 4: Use product rule: $$= -\frac{1}{x^2} \ln(\tan x) + \frac{1}{x} \cdot \frac{1}{\tan x} \cdot \sec^2 x$$ Step 5: Multiply both sides by $f(x)$: $$f'(x) = (\tan x)^{1/x} \left(-\frac{1}{x^2} \ln(\tan x) + \frac{\sec^2 x}{x \tan x} \right)$$