Re: [微分]
※ 引述《king911015 (早已放棄愛上你)》之銘言：
: lim 1
: 〔----- - (cotX)^2 〕
: x->0 X^2
: Find the slope of the line tangent to the graph of the equation
: sin(xy) = x^2 (cosy) at the point (2,π/2).
解法一
令 f(x,y) = sin(xy) - (x^2)(cosy)
dy fx (y)(cos(xy)) - (2x)(cosy)
---- = - ---- = - ----------------------------
dx fy (x)(cos(xy)) + (x^2)(siny)
dy | (π/2)(cos(π)) - (2)(2)(cos(π/2))
---- | = - -------------------------------------
dx |(2,π/2) (2)(cos(π)) + (4)(sin(π/2))
(π/2)(-1)
= - -------------------
(2)(-1) + (4)(1)
π/2 π
= ------ = ---
2 4
π
因此所求斜率為 ---
4
解法二
sin(xy) = (x^2)(cosy)
兩邊對 x 微分
dy dy
(cos(xy))(y + (x)(----)) = (2x)(cosy) + (x^2)(-siny)(----)
dx dx
dy
(----)((x)(cos(xy)) + (x^2)(siny)) = (2x)(cosy) - (y)(cos(xy))
dx
dy (2x)(cosy) - (y)(cos(xy))
---- = ----------------------------
dx (x)(cos(xy)) + (x^2)(siny)
dy | (4)(cos(π/2)) - (π/2)(cos(π))
---- | = ----------------------------------
dx |(2,π/2) (2)(cos(π)) + (4)(sin(π/2))
-(π/2)(-1)
= --------------
(2)(-1) + 4
π/2
= ------
2
π
= ---
4
π
因此所求斜率為 ---
4
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