3. Find the equation of the tangent to the curve \( y=\ln (3 x-2) \) at the point where \( x=1 \). 4. Show that if \( y=\frac{\cos x}{\sin x} \), then \( \frac{d y}{d x}=-\operatorname{cosec}^{2} x \).

To solve the problems, we will tackle them one at a time. ### Problem 3: Find the equation of the tangent to the curve \( y=\ln (3 x-2) \) at the point where \( x=1 \). 1. **Find the point on the curve at \( x=1 \)**: \[ y = \ln(3(1) - 2) = \ln(3 - 2) = \ln(1) = 0. \] So the point is \( (1, 0) \). 2. **Find the derivative \( \frac{dy}{dx} \)**: Using the chain rule, we differentiate \( y = \ln(3x - 2) \): \[ \frac{dy}{dx} = \frac{1}{3x - 2} \cdot \frac{d}{dx}(3x - 2) = \frac{1}{3x - 2} \cdot 3 = \frac{3}{3x - 2}. \] 3. **Evaluate the derivative at \( x=1 \)**: \[ \frac{dy}{dx} \bigg|_{x=1} = \frac{3}{3(1) - 2} = \frac{3}{3 - 2} = \frac{3}{1} = 3. \] 4. **Use the point-slope form of the line to find the equation of the tangent**: The point-slope form is given by: \[ y - y_1 = m(x - x_1), \] where \( m \) is the slope and \( (x_1, y_1) \) is the point on the curve. Here, \( m = 3 \) and \( (x_1, y_1) = (1, 0) \): \[ y - 0 = 3(x - 1). \] Simplifying this gives: \[ y = 3x - 3. \] Thus, the equation of the tangent line to the curve at the point where \( x=1 \) is: \[ \boxed{y = 3x - 3}. \] --- ### Problem 4: Show that if \( y=\frac{\cos x}{\sin x} \), then \( \frac{d y}{d x}=-\operatorname{cosec}^{2} x \). 1. **Rewrite \( y \)**: \[ y = \cot x = \frac{\cos x}{\sin x}. \] 2. **Differentiate \( y \) using the quotient rule**: The quotient rule states that if \( y = \frac{u}{v} \), then: \[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. \] Here, \( u = \cos x \) and \( v = \sin x \): - \( \frac{du}{dx} = -\sin x \) - \( \frac{dv}{dx} = \cos x \) Applying the quotient rule: \[ \frac{dy}{dx} = \frac{\sin x (-\sin x) - \cos x (\cos x)}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}. \] 3. **Use the Pythagorean identity**: Recall that \( \sin^2 x + \cos^2 x = 1 \): \[ \frac{dy}{dx} = \frac{-1}{\sin^2 x} = -\operatorname{cosec}^2 x. \] Thus, we have shown that: \[ \frac{dy}{dx} = -\operatorname{cosec}^2 x. \] Therefore, the result is: \[ \boxed{\frac{dy}{dx} = -\operatorname{cosec}^2 x}. \]