Use implicit differentiation to find an equation of the tangent line to the curve \( x^{2}-5 x y^{2}-7 y^{5}=1 \) at the point \( (2,-1) \). Use logarithmic differentiation to find the first derivative of the function \[ y=\frac{\ln (3 x)\left(2 x^{2}+5 x\right)^{3}}{\sqrt{4 x-3}} \]

To solve the two parts of your question, we will start with implicit differentiation for the first part and then use logarithmic differentiation for the second part. ### Part 1: Implicit Differentiation We need to find the equation of the tangent line to the curve given by: \[ x^{2} - 5xy^{2} - 7y^{5} = 1 \] at the point \( (2, -1) \). 1. **Differentiate both sides with respect to \( x \)**: \[ \frac{d}{dx}(x^{2}) - \frac{d}{dx}(5xy^{2}) - \frac{d}{dx}(7y^{5}) = \frac{d}{dx}(1) \] Using the product rule on \( 5xy^{2} \) and the chain rule on \( 7y^{5} \): \[ 2x - 5\left(y^{2} + 2xy\frac{dy}{dx}\right) - 7(5y^{4}\frac{dy}{dx}) = 0 \] 2. **Simplify the equation**: \[ 2x - 5y^{2} - 10xy\frac{dy}{dx} - 35y^{4}\frac{dy}{dx} = 0 \] 3. **Rearranging to solve for \( \frac{dy}{dx} \)**: \[ 2x - 5y^{2} = (10xy + 35y^{4})\frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{2x - 5y^{2}}{10xy + 35y^{4}} \] 4. **Evaluate \( \frac{dy}{dx} \) at the point \( (2, -1) \)**: Substituting \( x = 2 \) and \( y = -1 \): \[ \frac{dy}{dx} = \frac{2(2) - 5(-1)^{2}}{10(2)(-1) + 35(-1)^{4}} = \frac{4 - 5}{-20 + 35} = \frac{-1}{15} \] 5. **Equation of the tangent line**: Using the point-slope form of the line \( y - y_1 = m(x - x_1) \): \[ y - (-1) = -\frac{1}{15}(x - 2) \] This simplifies to: \[ y + 1 = -\frac{1}{15}x + \frac{2}{15} \] \[ y = -\frac{1}{15}x + \frac{2}{15} - 1 \] \[ y = -\frac{1}{15}x - \frac{13}{15} \] Thus, the equation of the tangent line is: \[ y = -\frac{1}{15}x - \frac{13}{15} \] ### Part 2: Logarithmic Differentiation We need to find the first derivative of the function: \[ y = \frac{\ln(3x)(2x^{2} + 5x)^{3}}{\sqrt{4x - 3}} \] 1. **Take the natural logarithm of both sides**: \[ \ln(y) = \ln\left(\ln(3x)(2x^{2} + 5x)^{3}\right) - \ln\left(\sqrt{4x - 3}\right) \] 2. **Using properties of logarithms**: \[ \ln(y) = \ln(\ln(3x)) + 3\ln(2x^{2} + 5x) - \frac{1}{2}\ln(4x - 3) \] 3. **Differentiate both sides with respect to \( x \)**: Using implicit differentiation: \[ \frac{1}{y}\frac{dy}{dx} = \frac{1}{\ln(3x)}\cdot\frac{1}{3x}\cdot 3 + 3\cdot\frac{1}{2x^{2} + 5x}\cdot(4x + 5) - \frac{1}{2(4x - 3)}\cdot 4 \] 4. **Multiply through by \( y \)** to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y\left(\frac{1}{3x} + \frac{3(4x + 5)}{2x^{2} + 5x} - \frac{2}{4x - 3}\right) \] 5. **Substituting back for \( y \)**: \[ \frac{dy}{dx} = \frac{\ln(3x)(2x^{2} + 5x)^{3}}{\sqrt{4x - 3}}\left(\frac{1}{3x} + \frac{3(4x + 5)}{2x^{2} + 5x} - \frac{2}{4x - 3}\right) \] This gives us the first derivative of the function using logarithmic differentiation.