作者LuisSantos (^______^)
看板trans_math
標題Re: [微分]
時間Tue Feb 6 12:59:56 2007
※ 引述《king911015 (早已放棄愛上你)》之銘言：
: lim 1
: 〔----- - (cotX)^2 〕
: x->0 X^2
1
lim (----- - (cotx)^2)
x→0 x^2
1 cosx
= lim (----- - (------)^2)
x→0 x^2 sinx
1 (cosx)^2
= lim (----- - ----------)
x→0 x^2 (sinx)^2
(sinx)^2 - (x^2)((cosx)^2)
= lim ----------------------------
x→0 (x^2)((sinx)^2)
((sinx) + (x)(cosx))((sinx) - (x)(cosx))
= lim -------------------------------------------
x→0 (sinx)(x^2)(sinx)
sinx + (x)(cosx) sinx - (x)(cosx)
= lim (------------------)(-------------------)
x→0 sinx (x^2)(sinx)
sinx + (x)(cosx) sinx - (x)(cosx)
= (lim ------------------)(lim ------------------)
x→0 sinx x→0 (x^2)(sinx)
cosx + cosx - (x)(sinx) cosx - cosx + (x)(sinx)
= (lim -------------------------)(lim --------------------------)
x→0 cosx x→0 (2x)(sinx) + (x^2)(cosx)
1 + 1 - 0 (x)(sinx)
= (-----------)(lim --------------------------)
1 x→0 (2x)(sinx) + (x^2)(cosx)
sinx + (x)(cosx)
= (2)(lim ----------------------------------------------------)
x→0 (2)(sinx) + (2x)(cosx) + (2x)(cosx) - (x^2)(sinx)
sinx + (x)(cosx)
= (2)(lim --------------------------------------)
x→0 (2)(sinx) + (4x)(cosx) - (x^2)(sinx)
cosx + cosx - (x)(sinx)
= (2)(lim ---------------------------------------------------)
x→0 (2)(cosx) + (4)(cosx) - (2x)(sinx) - (x^2)(cosx)
(2)(cosx) - (x)(sinx)
= (2)(lim --------------------------------------)
x→0 (6)(cosx) - (2x)(sinx) - (x^2)(cosx)
2 - 0
= (2)(-----------)
6 - 0 - 0
2 2
= (2)(---) = ---
6 3
: Find the slope of the line tangent to the graph of the equation
: sin(xy) = x^2 (cosy) at the point (2,π/2).
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