請問這題怎麼算? 微分高手請進

f(x,y) = (x^2+y)^{1/5}
a=5, b=7 則 f(a,b) = (5^2+7)^{1/5} = 2.
df(x,y) = f_x(a,b) dx + f_y(a,b) dy
f_x(x,y) = (2/5)x(x^2+y)^{-4/5}, 故 f_x(a,b) = (2/5)(5)(1/16) = 1/8
f_y(x,y) = (1/5)(x^2+y)^{-4/5}, 故 f_y(a,b) = (1/5)(1/16) = 1/80
dx = -0.05, dy = 0.02
f(x,y) ≒ f(a,b) + df(x,y)
故
f(4.95,7.02) ≒ f(5,7) + f_x(5.7) dx + f_y(5,7) dy
= 2 +(1/8)(-0.05) + (1/80)(0.02)
= 2 - 0.00625 + 0.00025 = 1.94
(4.95^2+7.02)^{1/5} ≒ 1.94

2011-06-02 02:13:21 補充：
f(4.95,7.02) ≒ 2 - 0.00625 + 0.00025 = 1.994
(4.95^2+7.02)^{1/5} ≒ 1.994

( 4.95^2 + 7.02 ) ^ (1/5)

=(24.502+7.02)^(1/5)

=(31.5225)^(1/5)

取 log(31.5225)^(1/5)

=(1/5)*log(31.5225)

=(1/5)*(1+log3.15225) 查對數表得到

=(1/5)*(1+0.4983)

=0.29966

log x=0.29966

查對數表得到

x=1.994---

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