Derivatives of parametric functions
The position of a moving car is given by two clocks: x at time t and y at time t. The slope of its path is found by dividing how fast y changes by how fast x changes.
If x = f(t) and y = g(t), then dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
Formula: dy/dx = (dy/dt) / (dx/dt)
Rule 1: Find dy/dx — Differentiate each with respect to t and divide.
- Easy: x = 2t, y = t² gives dy/dx = 2t / 2 = t.
- Medium: x = t², y = t³ gives dy/dx = 3t² / 2t = 3t/2.
- Exam-type: x = a cos θ, y = a sin θ gives dy/dx = −cot θ.
Rule 2: Slope at a given t — Substitute t after dividing.
- Easy: x = t + 1, y = t² − t at t = 2: dy/dx = (2t − 1) / 1 = 3.
- Medium: x = 3t, y = t³ at t = 1: dy/dx = 3t² / 3 = 1.
- Exam-type: x = 2 cos t, y = 3 sin t at t = π/4: dy/dx = −(3/2) cot t = −3/2.
The parameter t cancels in the sense that the final slope may still contain t.
Writing dy/dx = (dx/dt) / (dy/dt), upside down.
"y over x" in the same order: dy/dt over dx/dt.
Practice questions
- 1Find dy/dx if x = 3t, y = t².
- 2Find dy/dx if x = t², y = 2t.
- 3Find dy/dx if x = t³, y = t².
- 4Find dy/dx if x = 4 cos t, y = 4 sin t.
- 5Find dy/dx if x = t + 1/t, y = t − 1/t.
- 6Find dy/dx if x = eᵗ, y = e⁻ᵗ.
- 7Find dy/dx at t = 1 if x = 2t², y = 4t.
- 8Find dy/dx if x = sin t, y = cos 2t.
Show answers (try them yourself first)
- 12t / 3
- 22 / 2t = 1/t
- 32t / 3t² = 2 / (3t)
- 4−4 sin t / 4 cos t = −cot t
- 5(1 + 1/t²) / (1 − 1/t²) = (t² + 1) / (t² − 1)
- 6−e⁻ᵗ / eᵗ = −e⁻²ᵗ
- 74 / (4t) = 1 at t = 1
- 8−2 sin 2t / cos t = −4 sin t
Second order derivative
Speed is the derivative of distance, and acceleration is the derivative of speed. Taking a derivative twice gives the second order derivative.
The second order derivative of y = f(x) is f″(x) = d/dx (f′(x)), also written d²y/dx² or y″. For a parametric function, d²y/dx² = [d/dt (dy/dx)] / (dx/dt).
Formula: y″ = d/dx (dy/dx)
Rule 1: Differentiate twice — Differentiate once, then again.
- Easy: y = x⁴ gives y′ = 4x³ and y″ = 12x².
- Medium: y = sin 2x gives y′ = 2 cos 2x and y″ = −4 sin 2x.
- Exam-type: y = x³ + 3x² − 5x gives y′ = 3x² + 6x − 5 and y″ = 6x + 6.
Rule 2: Prove a relation — Compute y′ and y″ and substitute.
- Easy: y = e^(2x): y″ = 4e^(2x) = 4y, so y″ − 4y = 0.
- Medium: y = sin x + cos x: y″ = −sin x − cos x = −y, so y″ + y = 0.
- Exam-type: x = t², y = t³ gives dy/dx = 3t/2, then d²y/dx² = (3/2) / (2t) = 3 / (4t).
The second derivative of a function tells how fast its slope is changing.
Writing d²y/dx² as (dy/dx)²; it is the derivative of the derivative.
Derive, then derive again.
Practice questions
- 1Find d²y/dx² for y = x⁵.
- 2Find d²y/dx² for y = sin 3x.
- 3Find d²y/dx² for y = e⁻ˣ.
- 4Find d²y/dx² for y = x ln x.
- 5Show that y = e^(3x) satisfies y″ − 9y = 0.
- 6Show that y = 3 sin x + 4 cos x satisfies y″ + y = 0.
- 7Find d²y/dx² if x = 2t, y = t³.
- 8Find y″ for y = x² eˣ.
Show answers (try them yourself first)
- 120x³
- 2−9 sin 3x
- 3e⁻ˣ
- 4y′ = ln x + 1, so y″ = 1/x
- 5y″ = 9e^(3x) = 9y
- 6y″ = −3 sin x − 4 cos x = −y
- 7dy/dx = 3t²/2; derivative with respect to t is 3t; divide by dx/dt = 2 to get 3t/2
- 8y′ = eˣ (x² + 2x), y″ = eˣ (x² + 4x + 2)