n阶求导……
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发布时间：2022-07-20 16:28
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热心网友
时间：2023-10-22 22:46
y'' = 2cos2x = -2cos(2x+π), y''' = -2^2sin2x = -2^2cos(2x+3π/2),
y^(4) = -2^3cos2x = -2^3cos(2x+2π), ......, y^(n) = -2^(n-1)cos(2x+nπ/2).
2. y = ln(x+1), y' = 1/(x+1) = (x+1)^(-1),
y'' = (-1)(x+1)^(-2), y''' = (-1)(-2)(x+1)^(-3), ......,
y^(n) = (-1)^(n-1) (n-1)!/(x+1)^n
3. y = 1/(x^2-1) = (1/2)[1/(x-1)-1/(x+1)] = (1/2)[(x-1)^(-1)-(x+1)^(-1)] .
y' = (1/2)[(-1)(x-1)^(-2) - (-1)(x+1)^(-2)]
y'' = (1/2)[(-1)(-2)(x-1)^(-3) - (-1)(-2)(x+1)^(-3)]
............
y^(n) = (1/2)(-1)^n n! [1/(x-1)^(n+1) - 1/(x+1)^(n+1)]
4. y = x(x+1)(x+2)......(x+n) = x[x^n + (1/2)n(n+1)x^(n-1) + ......]
= x^(n+1) + (1/2)n(n+1)x^n + ......
y' = (n+1)x^n + (1/2)n(n+1)nx^(n-1) + ......
y'' = (n+1)nx^(n-1) + (1/2)n(n+1)n(n-1)x^(n-2) + ......
......
y^(n) = (n+1)! x + (1/2)n(n+1) n!
