Continuity and Differentiability — Practice Quiz

-sin x / (1 + cos y)

Step 1: Differentiate both sides of y + sin y = cos x with respect to x. Step 2: d/dx(y) + d/dx(sin y) = d/dx(cos x) gives dy/dx + cos y · dy/dx = -sin x. Step 3: Factor out dy/dx: dy/dx(1 + cos y) = -sin x. Step 4: Therefore dy/dx = -sin x / (1 + cos y), valid when y ≠ (2n+1)π. Final: The answer is -sin x/(1 + cos y). Common mistake: Forgetting to apply chain rule when differentiating sin y.

1 / √(1 - x²)

Step 1: Let y = sin⁻¹ x, so sin y = x. Step 2: Differentiate both sides: cos y · dy/dx = 1, giving dy/dx = 1/cos y. Step 3: Since cos²y = 1 - sin²y = 1 - x², and cos y > 0 for y ∈ (-π/2, π/2), cos y = √(1 - x²). Step 4: Therefore dy/dx = 1/√(1 - x²), valid for x ∈ (-1, 1). Final: The answer is 1/√(1 - x²). Common mistake: Confusing this with the derivative of cos⁻¹ x which has a negative sign.

1/t

Step 1: We have parametric equations x = at², y = 2at. Step 2: Find dx/dt = 2at and dy/dt = 2a. Step 3: Apply the parametric differentiation formula: dy/dx = (dy/dt) / (dx/dt). Step 4: dy/dx = 2a / (2at) = 1/t. Final: The answer is 1/t. This represents the slope of the tangent to the parabola. Common mistake: Some students incorrectly compute dx/dy instead of dy/dx.

6x + 2sec²x tan x

Step 1: Start with y = x³ + tan x. Step 2: Find first derivative: dy/dx = 3x² + sec²x. Step 3: Differentiate again to get d²y/dx²: d/dx(3x²) = 6x. Step 4: d/dx(sec²x) = 2sec x · d/dx(sec x) = 2sec x · sec x tan x = 2sec²x tan x. Final: Therefore d²y/dx² = 6x + 2sec²x tan x. Common mistake: Forgetting to apply chain rule when differentiating sec²x.