推 sioumi:第一行的fxy...應該是fyx 59.113.114.76 01/31 13:29
※ 編輯: LuisSantos 來自: 61.66.173.21 (02/01 00:47)
※ 引述《tiyico (宏)》之銘言：
: xy(x^2-y^2)
: { ------------- (x,y) =/= (0,0)
: 01. f(x,y) ={ x^2+y^2
: {
: 0 (x,y) = (0,0)
: then f (0,0) = ? f (0,0) = ?
: xy yx
δ(fy)
f (0,0) = --------(0,0)
yx δx
fy(Δx , 0) - fy(0,0)
= lim ----------------------
Δx->0 Δx
其中
fy(Δx,0)
f(Δx,Δy) - f(Δx,0)
= lim ----------------------
Δy->0 Δy
((Δx)(Δy)((Δx)^2 - (Δy)^2))/((Δx)^2 + (Δy)^2) - 0
= lim -------------------------------------------------------- = Δx
Δy->0 Δy
f(0,Δy) - f(0,0) 0
而 fy(0,0) = lim ----------------- = lim ----- = 0
Δy->0 Δy Δy->0 Δy
Δx - 0
則 fyx(0,0) = lim --------- = 1
Δx->0 Δx
δ(fx) fx(0,Δy) - fx(0,0)
fxy(0,0) = ------(0,0) = lim -------------------
δy Δy->0 Δy
其中
fx(0,Δy)
f(Δx,Δy) - f(0,Δy)
= lim ---------------------
Δx->0 Δx
((Δx)(Δy)((Δx)^2 - (Δy)^2))/((Δx)^2 + (Δy)^2) - 0
= lim -------------------------------------------------------- = -Δy
Δx->0 Δx
f(Δx,0) - f(0,0) 0
而fx(0,0) = lim ------------------ = lim ----- = 0
Δx->0 Δx Δx->0 Δx
-Δy - 0
則fxy(0,0) = lim ---------- = -1
Δy->0 Δy
: x^2 y^2 z^2
: 02. Let R be the region inside the ellipsiod ----- + ----- + ----- = 1
: a^2 b^2 c^2
: (a,b,c >0)
: and above the plane z = b-y, then the volumn of the region R is ??
: x^2˙sin(1/x)
: 03. lim ------------- (我算0但是不確定耶..)
: x->0 tanx
: 1 sin(πx^2) 1
: 04. prove 0 < ∫ ---------- dx < ---ln2
: -1/√2 x 2
: 沒頭緒
: 感激賜教了
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