1. **Problem:** Find the derivative of $f(x) = e^{3x}$. 2. **Formula:** The derivative of $e^{u(x)}$ is $e^{u(x)} \cdot u'(x)$. 3. **Step:** Here, $u(x) = 3x$, so $u'(x) = 3$. 4. **Derivative:** $$f'(x) = e^{3x} \cdot 3 = 3e^{3x}$$ --- 1. **Problem:** Find the derivative of $f(x) = e^{2x+5}$. 2. **Formula:** The derivative of $e^{u(x)}$ is $e^{u(x)} \cdot u'(x)$. 3. **Step:** Here, $u(x) = 2x + 5$, so $u'(x) = 2$. 4. **Derivative:** $$f'(x) = e^{2x+5} \cdot 2 = 2e^{2x+5}$$ --- 1. **Problem:** Find the derivative of $f(x) = \log_{10}(3x) + 7$. 2. **Formula:** The derivative of $\log_a(x)$ is $\frac{1}{x \ln(a)}$. 3. **Step:** For $\log_{10}(3x)$, use chain rule: derivative is $\frac{1}{3x \ln(10)} \cdot 3 = \frac{3}{3x \ln(10)}$. 4. **Simplify:** $$\frac{3}{3x \ln(10)} = \frac{1}{x \ln(10)}$$ 5. **Derivative:** The constant 7 derivative is 0, so $$f'(x) = \frac{1}{x \ln(10)}$$ --- 1. **Problem:** Find the derivative of $f(x) = \ln\left(\frac{3}{\sqrt{x^3 + 3x + 4}}\right)$. 2. **Rewrite:** $$f(x) = \ln(3) - \ln\left((x^3 + 3x + 4)^{1/2}\right) = \ln(3) - \frac{1}{2} \ln(x^3 + 3x + 4)$$ 3. **Derivative:** Derivative of $\ln(3)$ is 0. 4. **Use chain rule:** $$f'(x) = -\frac{1}{2} \cdot \frac{1}{x^3 + 3x + 4} \cdot (3x^2 + 3)$$ 5. **Simplify:** $$f'(x) = -\frac{3x^2 + 3}{2(x^3 + 3x + 4)}$$ --- 1. **Problem:** Find the derivative of $f(x) = \ln\left(e^{4x^3 + 3x^2 + 2}\right)$. 2. **Simplify:** $$f(x) = 4x^3 + 3x^2 + 2$$ because $\ln(e^u) = u$. 3. **Derivative:** $$f'(x) = 12x^2 + 6x$$ --- **Final answers:** 1. $$f'(x) = 3e^{3x}$$ 2. $$f'(x) = 2e^{2x+5}$$ 3. $$f'(x) = \frac{1}{x \ln(10)}$$ 4. $$f'(x) = -\frac{3x^2 + 3}{2(x^3 + 3x + 4)}$$ 5. $$f'(x) = 12x^2 + 6x$$