If y = √(2x + 1), then differentiating both sides implicitly with respect to t gives
dy/dt = 1/2 • 1/√(2x + 1) • 2 • dx/dt = 1/√(2x + 1) • dx/dt
(a) If dx/dt = 9 and x = 4, then
dy/dt = 1/√(2•4 + 1) • 9
dy/dt = 1/√(8 + 1) • 9
dy/dt = 1/√9 • 9
dy/dt = 9/3
dy/dt = 3
(b) If dy/dt = 3 and x = 40, then
3 = 1/√(2•40 + 1) • dx/dt
3 = 1/√(80 + 1) • dx/dt
3 = 1/√81 • dx/dt
3 = 1/9 • dx/dt
dx/dt = 27
