- 2068 Chaitra
Derivatives of y^(1/m) + y^(−1/m) = 2x
Short answer
Let t = y^(1/m); then t+1/t=2x ⇒ y = (x+√(x²−1))^m. Differentiate: y₁ = m y/√(x²−1) ⇒ (x²−1)y₁² = m²y². This yields (x²−1)y₂ + xy₁ − m²y = 0. Differentiating n times by Leibniz gives (x²−1)y_{n+2} + (2n+1)xy_{n+1} + (n²−m²)y_n = 0.
If y^(1/m) + y^(−1/m) = 2x, show that: a) (x²−1)y₂ + xy₁ − m²y = 0, b) (x²−1)y_{n+2} + (2n+1)xy_{n+1} + (n²−m²)y_n = 0.
Given y^(1/m) + y^(−1/m) = 2x. Put t = y^(1/m); then t + 1/t = 2x, so t² − 2xt + 1 = 0 and t = x ± √(x² − 1). Hence
y = (x + √(x² − 1))^m (the other sign gives the same equations below).
Differentiate: y₁ = m (x + √(x²−1))^(m−1) · [1 + x/√(x²−1)] = m y / √(x² − 1).
So (x² − 1) y₁² = m² y² …(1)
(a)
Differentiate (1): 2(x² − 1) y₁y₂ + 2x y₁² = 2m² y y₁. Divide by 2y₁:
(x² − 1) y₂ + x y₁ − m² y = 0. Proved. …(2)
(b)
Differentiate (2) n times by Leibnitz's theorem:
(x²−1)yₙ₊₂ + n(2x)yₙ₊₁ + [n(n−1)/2](2)yₙ + x yₙ₊₁ + n yₙ − m² yₙ = 0
The coefficient of yₙ is n(n−1) + n − m² = n² − m².
(x² − 1) yₙ₊₂ + (2n+1) x yₙ₊₁ + (n² − m²) yₙ = 0. Proved.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Derivatives of y = (sin⁻¹x)² · Mathematics I, chapter 1
- Derivatives of y = sin(m sin⁻¹x) · Mathematics I, chapter 1
- Leibnitz theorem: y = sin(m sin⁻¹x) · Mathematics I, chapter 1 · asked 2×
- Leibnitz theorem: y = e^(a tan⁻¹x) · Mathematics I, chapter 1 · asked 2×
- Leibnitz theorem: y = log(x+√(a²+x²)) · Mathematics I, chapter 1 · asked 3×
- Leibnitz theorem: y = (sin⁻¹x)² · Mathematics I, chapter 1
- Verify Stokes' theorem F=(2x−y)i−yz²j−y²zk · Mathematics III, chapter 2 · asked 3×
- pdf k(5+2x): find k, mean, variance · Probability and Statistics, chapter 3 · asked 2×
- Derivatives at t=0+: 1 H, 1 F, 10 V · Electrical Circuit Theory, chapter 2 · asked 2×
- Verify pdf 2e^(−2x); P(1<x<3); mean · Probability and Statistics, chapter 3