微分 ＊
EX 2.
4
√
16.01×
5
√
32.1
設z=f
(
x , y
)
=x
1
4
y
1
5
⇒dz=1
4x
−3
4
y
1
5
dx+1
5x
1
4
y
−4
5
dy
(
x , y , dx , dy
)
=
(
16,32,0.01 ,0.1
)
代入上式得dz =1
4∙1
8∙2∙0.01+1
5∙2∙1
16 ∙0.1=0.003125
⇒所求 ≒f
(
16,32
)
+0.003125=4.003125
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math exercise and differentia and some integral
Typology: Exercises
2019/2020
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EX 2.
4
√16.01 ×^
5
設 z = f ( x , y )= x 1 (^4) y 1 (^5) ⇒dz = 1 4 x − 3 (^4) y 1 (^5) dx + 1 5 x 1 (^4) y − 4 (^5) dy ( x , y , dx , dy )=( 16,32,0.01 , 0.1) 代入上式得 dz =
⇒ 所求 ≒f ( 16,32) +0.003125=4.