Second order derivatives

If y = f(x) and f′ is itself differentiable, its derivative is the second derivative, written d²y/dx², f″(x), D²y, y″ or y₂.

Worked example: y = x³ + tan x. y′ = 3x² + sec²x, then y″ = 6x + 2 sec²x tan x.

Worked example: y = A sin x + B cos x. y′ = A cos x − B sin x and y″ = −A sin x − B cos x = −y, so y″ + y = 0 for any constants A and B.

Worked example: y = 3e²ˣ + 2e³ˣ. Here y′ = 6e²ˣ + 6e³ˣ and y″ = 12e²ˣ + 18e³ˣ, which gives y″ − 5y′ + 6y = 0.

Worked example: y = sin⁻¹x. Then √(1 − x²) y′ = 1; differentiating again and clearing the root gives (1 − x²)y″ − xy′ = 0.

Strategy for 'prove that' questions: first write y′ in a form free of roots and fractions, then differentiate again. It avoids a messy quotient rule.

In parametric form, y″ is not (d²y/dt²)/(d²x/dt²). Differentiate dy/dx with respect to t, then divide by dx/dt.