Solution

(a) \[ y’ = \frac{1/2}{1 + (x/2)^2} = \frac{2}{4 + x^2} \] At \( x = 2 \), slope \( m = \frac{1}{4} \) \[ y – \frac{\pi}{4} = \frac{1}{4}(x – 2) \]

(b) Using product rule: \[ y’ = 4\arccos(x-1) – \frac{4x}{\sqrt{2x – x^2}} \] At \( x = 1 \), slope \( m = 4\pi – 4 \) \[ y – 2\pi = (4\pi – 4)(x – 1) \]

(c) Using implicit differentiation: \[ 2x + \arctan y + \frac{x}{1 + y^2}y’ = y’ \] At \( \left( -\frac{\pi}{4}, 1 \right) \), solve for \( y’ \) to get slope \( m \)