62.求下列各式的近似值：

(1)(0.95)^1/5
设 y = f(x) = x^(1/5)
选 xo = 1, Δx = 0.95 - 1 = -0.05
f(xo) = f(1) = 1
f'(x) = dy/dx = (1/5)x^(-4/5)
f'(xo)= f'(1) = 1/5
∴ (0.95)^1/5 = f(xo) + f'(xo)Δx
= 1 + (1/5)×(-0.05)
= 0.99
[对比：准确值=0.989794，误差0.02%]

因为本题题意不清，另一种可能是：

(1)(√0.95)^1/5 = (0.95)^1/10
设 y = f(x) = x^(1/10)
选 xo = 1, Δx = 0.95 - 1 = -0.05
f(xo) = f(1) = 1
f'(x) = dy/dx = (1/10)x^(-9/10)
f'(xo)= f'(1) = 1/10
∴ (√0.95)^1/5 = f(xo) + f'(xo)Δx
= 1 + (1/10)×(-0.05)
= 0.995
[对比：准确值=0.994884，误差0.01%]

(3)ln1.01
设 y = f(x) = lnx
选 xo = 1, Δx = 1.01 - 1 = 0.01
f(xo) = f(1) = 0
f'(x) = dy/dx = 1/x
f'(xo)= f'(1) = 1
∴ ln1.01 = f(xo) + f'(xo)Δx
= 0 + 1×(0.01)
= 0.01
[对比：准确值=0.009950,误差0.50%]

(6) arctan1.02
设 y = f(x) = arctanx
选 xo = 1, Δx = 1.02 - 1 = 0.02
f(xo) = f(1) = π/4
f'(x) = dy/dx = 1/(1+x²)
f'(xo)= f'(1) = 1/2
∴ arctan1.02 = f(xo) + f'(xo)Δx
= π/4 + ½×(0.02)
= 0.795398
[对比：准确值=0.795299,误差0.01%]

说明：考试时，不需要提供准确值，也不需要误差。只要算出来即可。