1. **Statement of the problem:** Calculate the derivatives of the following functions: $$f_1(x) = 2x \sqrt{-3x + 2}$$ $$f_2(x) = (-x + 4) e^{-x}$$ $$f_3(x) = \frac{\ln(x^2 + 1)}{x^2 + 1}$$ $$f_4(x) = -x + 7 + 6 \ln(2x + 1) - 6 \ln(2x + 2)$$ $$f_5(x) = \frac{5^x}{5^{2x - 1}}$$ 2. **Formulas and rules used:** - Derivative of product: $\frac{d}{dx}[uv] = u'v + uv'$ - Derivative of quotient: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$ - Derivative of $\sqrt{g(x)} = \frac{g'(x)}{2\sqrt{g(x)}}$ - Derivative of $e^{g(x)} = e^{g(x)} g'(x)$ - Derivative of $\ln(g(x)) = \frac{g'(x)}{g(x)}$ - Derivative of $a^x = a^x \ln(a)$ 3. **Step-by-step solutions:** **For $f_1(x) = 2x \sqrt{-3x + 2}$:** - Let $u = 2x$, $v = \sqrt{-3x + 2} = (-3x + 2)^{1/2}$ - $u' = 2$ - $v' = \frac{1}{2}(-3x + 2)^{-1/2} \cdot (-3) = -\frac{3}{2\sqrt{-3x + 2}}$ - Using product rule: $$f_1'(x) = u'v + uv' = 2 \sqrt{-3x + 2} + 2x \left(-\frac{3}{2\sqrt{-3x + 2}}\right) = 2 \sqrt{-3x + 2} - \frac{3x}{\sqrt{-3x + 2}}$$ **For $f_2(x) = (-x + 4) e^{-x}$:** - Let $u = -x + 4$, $v = e^{-x}$ - $u' = -1$ - $v' = -e^{-x}$ - Using product rule: $$f_2'(x) = u'v + uv' = (-1) e^{-x} + (-x + 4)(-e^{-x}) = -e^{-x} + (x - 4) e^{-x} = (x - 5) e^{-x}$$ **For $f_3(x) = \frac{\ln(x^2 + 1)}{x^2 + 1}$:** - Let $u = \ln(x^2 + 1)$, $v = x^2 + 1$ - $u' = \frac{2x}{x^2 + 1}$ - $v' = 2x$ - Using quotient rule: $$f_3'(x) = \frac{u'v - uv'}{v^2} = \frac{\frac{2x}{x^2 + 1} (x^2 + 1) - \ln(x^2 + 1) (2x)}{(x^2 + 1)^2} = \frac{2x - 2x \ln(x^2 + 1)}{(x^2 + 1)^2} = \frac{2x (1 - \ln(x^2 + 1))}{(x^2 + 1)^2}$$ **For $f_4(x) = -x + 7 + 6 \ln(2x + 1) - 6 \ln(2x + 2)$:** - Derivative of $-x$ is $-1$ - Derivative of constant $7$ is $0$ - Derivative of $6 \ln(2x + 1)$ is $6 \cdot \frac{2}{2x + 1} = \frac{12}{2x + 1}$ - Derivative of $-6 \ln(2x + 2)$ is $-6 \cdot \frac{2}{2x + 2} = -\frac{12}{2x + 2}$ - Summing up: $$f_4'(x) = -1 + \frac{12}{2x + 1} - \frac{12}{2x + 2}$$ **For $f_5(x) = \frac{5^x}{5^{2x - 1}}$:** - Simplify first: $$f_5(x) = 5^{x - (2x - 1)} = 5^{x - 2x + 1} = 5^{1 - x}$$ - Derivative: $$f_5'(x) = 5^{1 - x} \cdot \ln(5) \cdot (-1) = - \ln(5) \cdot 5^{1 - x}$$ 4. **Final answers:** $$f_1'(x) = 2 \sqrt{-3x + 2} - \frac{3x}{\sqrt{-3x + 2}}$$ $$f_2'(x) = (x - 5) e^{-x}$$ $$f_3'(x) = \frac{2x (1 - \ln(x^2 + 1))}{(x^2 + 1)^2}$$ $$f_4'(x) = -1 + \frac{12}{2x + 1} - \frac{12}{2x + 2}$$ $$f_5'(x) = - \ln(5) \cdot 5^{1 - x}$$