Re: [微積] 一個隱含數問題
※ 引述《Jolinfuns (嘎嘎)》之銘言：
: 2
: (x+y) dy
: 題目：---------- = 3x . Find ----
: x-y dx
: 問題：
: 用Quotient Rule(除法微分)去微分
: 與先把x-y乘到3x那邊來，再做微分
: 答案不一樣
: 但代值進去是一樣的
: 不知道是為什麼會這樣？
Ans1:
2
2(x+y)(1+dy/dx)(x-y)-(x+y) (1-dy/dx)
------------------------------------- = 3
2
(x-y)
2 2 2 2 2 2 2
=> 2(x - y )+2(x -y )dy/dx-(x+y) +(x+y) dy/dx = 3(x-y)
2 2 2 2 2 2 2
=>[2(x -y )+(x+y) ]dy/dx = 3(x-y) -2(x -y )+(x+y)
2 2 2 2 2
=>(3x -y +2xy)dy/dx = 3(x-y) -x +3y +2xy
2 2
2x +6y -4xy
=>dy/dx = --------------
2 2
3x -y +2xy
Ans2:
2
(x+y) = 3x(x-y)
2 2 2
=> x +2xy+y = 3x -3xy
=>2x+2y+(2x)dy/dx+(2y)dy/dx = 6x-3y-(3x)dy/dx
=>dy/dx(5x+2y)=4x-5y
4x-5y
=> dy/dx = -------
5x+2y
--
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