Second order derivatives
The second derivative d²y/dx² is the derivative of dy/dx. Board questions often ask you to prove a relation between y, y₁ and y₂.
What it is
If y = f(x), then dy/dx is again a function of x. Differentiating it once more gives the second order derivative, written d²y/dx², y₂, y″ or f″(x). In motion, if s is distance and t is time, ds/dt is velocity and d²s/dt² is acceleration.
Example 1: direct
Find d²y/dx² if y = x⁴ + tan x.
- dy/dx = 4x³ + sec²x
- Chain rule: d/dx (sec²x) = 2 sec x × sec x tan x = 2 sec²x tan x
- d²y/dx² = 12x² + 2 sec²x tan x
d²y/dx² = 12x² + 2 sec²x tan x
Example 2: show a relation
If y = e^(2x)(a + bx), where a and b are constants, show that y₂ − 4y₁ + 4y = 0.
- y₁ = 2e^(2x)(a + bx) + b e^(2x) = 2y + b e^(2x)
- So b e^(2x) = y₁ − 2y ... (1)
- Differentiate y₁ = 2y + b e^(2x): y₂ = 2y₁ + 2b e^(2x)
- Use (1): y₂ = 2y₁ + 2(y₁ − 2y) = 4y₁ − 4y
- So y₂ − 4y₁ + 4y = 0
Shown: y₂ − 4y₁ + 4y = 0
Example 3: clear the fraction first
If y = sin(log x), x > 0, show that x²y₂ + xy₁ + y = 0.
- y₁ = cos(log x) × (1/x)
- Clear the fraction: xy₁ = cos(log x)
- Differentiate both sides: y₁ + xy₂ = −sin(log x) × (1/x) = −y/x
- Multiply by x: xy₁ + x²y₂ = −y
x²y₂ + xy₁ + y = 0
Tips for 'show that' questions
- Find y₁, then multiply out to remove any fraction or square root.
- If y₁ contains the original function, write it in terms of y; the second step becomes shorter.
- Differentiate the cleared equation using the product rule.
- Rearrange to match the required form exactly, then write 'Hence proved'.
- d²y/dx² ≠ (dy/dx)². Never square the first derivative.
A speedometer shows ds/dt. The push you feel when a bus brakes suddenly comes from d²s/dt², the acceleration. Railway engineers keep it gentle on curves so that passengers stay comfortable.
1 marks: “If y = e^(3x), find d²y/dx².” dy/dx = 3e^(3x), so d²y/dx² = 9e^(3x).
3 marks: “If y = x sin x, show that x²y₂ − 2xy₁ + (x² + 2)y = 0.” y₁ = sin x + x cos x and y₂ = 2 cos x − x sin x. Then x²y₂ − 2xy₁ + (x² + 2)y = (2x² cos x − x³ sin x) − (2x sin x + 2x² cos x) + (x³ sin x + 2x sin x) = 0.
3 marks: “If x = 2(θ + sin θ) and y = 2(1 − cos θ), find d²y/dx² at θ = π/2.” dy/dx = tan(θ/2). d²y/dx² = (1/2) sec²(θ/2) ÷ 2(1 + cos θ). Since 1 + cos θ = 2 cos²(θ/2), this is sec⁴(θ/2)/8. At θ = π/2, sec²(π/4) = 2, so d²y/dx² = 4/8 = 1/2.
Second derivative: differentiate twice, never square.