Question

高数导数与极限

高数导数与极限求各阶导数

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Answer 1

如图所示

Answer 2

(1) y = x(lnx)^2, y' = (lnx)^2 + 2lnx,
y'' = 2lnx/x + 2/x, y''' = 2(1-lnx)/x^2 - 2/x^2 = -2lnx/x^2
(2) y = sin√x, y' = (1/2)x^(-1/2)cos√x,
y'' = (1/2)[(-1/2)x^(-3/2)cos√x - (1/2)x^(-1)sin√x]
= (-1/4)[x^(-3/2)cos√x + x^(-1)sin√x]
y''' = (-1/4)[(-3/2)x^(-5/2)cos√x - (1/2)x^(-2)sin√x
- x^(-2)sin√x + (1/2)x^(-3/2)cos√x]
= (-1/4)[(-3/2)x^(-5/2)cos√x - (3/2)x^(-2)sin√x + (1/2)x^(-3/2)cos√x]
= (1/8)[(3-x)cos√x/x^(5/2) + 3sin√x/x^2]
(4) y = √cosx, y' = (-1/2)(cosx)^(-1/2)sinx,
y'' = (-1/2)[(1/2)(cosx)^(-3/2)(sinx)^2 + (cosx)^(1/2)]
y''' = (-1/2)[(1/2)(3/2)(cosx)^(-5/2)(sinx)^3 + (cosx)^(-1/2)sinx - (1/2)(cosx)^(-1/2)sinx]
= (-1/2)[(3/4)(cosx)^(-5/2)(sinx)^3 + (1/2)(cosx)^(-1/2)sinx]
= (-1/8)[3(cosx)^(-5/2)(sinx)^3 + 2(cosx)^(-1/2)sinx]

y^(4) = (-1/8)[3(5/2)(cosx)^(-7/2)(sinx)^4 + 9(cosx)^(-3/2)(sinx)^2
+ 2(cosx)^(-3/2)(sinx)^2 + 2(cosx)^(1/2)]
= (-1/8)√cosx[(15/2)(cosx)^(-4)(sinx)^4 + 11(cosx)^(-2)(sinx)^2 + 2]
= (-1/8)√cosx[(15/2)(tanx)^4 + 11(tanx)^2 + 2]