函數的導數

函數的導數

函數的導數

According to the definition of derivative , f'(2) = lim f(x) - f(2) / x - 2 at x → 2 . 
帶入即可.

f'(3) = lim [ f(x) - f(3) ] / (x -3)  = lim [(x+1)^1/2 - 2] / x-3 = lim [ (x+1) - 4 ] / (x-3)[(x+1)^1/2 +2] = lim (x-3)/(x-3)[(x+1)^1/2 + 2]  = lim 1/(x+1)^1/2 +2 = 1/2+2= 1/4

f'(0) = lim [ [ x(x+1)(x+2)(x+3)/(1-x)(2-x)(3-x) ] - 0 ] / (x - 0) = lim (x+1)(x+2)(x+3)/(1-x)(2-x)(3-x) = 6 / 6 = 1

見絕對值必討論其中的式子 , |2x-4| .
2x- 4 < 0 , 2x < 4 , x < 2 故 x 趨近 3 , |2x-4| = 2x - 4 .
f'(3) = lim  [ x|2x-4| / |x| -2  - f(3) ] / (x - 3) = lim [x(2x-4) / x -2 - 6] / x -3  = lim 2x^2 - 4x / x -2 - 6 /x-3 = lim 2x^2 - 4x - 6x + 12 / (x-2)(x-3) = lim 2x^2 -10x +12 /(x-2)(x-3) = lim 2(x-3)(x-2)/(x-3)(x-2) = 2

x^2/25 + y^2/16 = 1 , 令其切線為 y = mx + b , 將其帶入,得
x^2/25 + (mx+b)^2 /16  =1
通分, 16x^2 + 25(mx+b)^2  = 25*16
(4x)^2 + [5(mx+b)]^2 = 20^2
16x^2 + 25(m^2x^2+2mbx+b^2) = 400
(16+25m^2)x^2 + (50mb)x + 25b^2 - 400 = 0

因為切線僅與橢圓有一交點,所以上方程式必有重根.
所以以根的判別式 2500(mb)^2 - 4(16+25m^2)(25b^2 -400) = 0
100(mb)^2 - 4(16+25m^2)(b^2-16) = 0
25(mb)^2 - (16+25m^2)(b^2-16) = 0
此法為麻煩.

假設該線的切點為(a,b) , y' = 2x + 1 . 故 m = 2a+1.
b = a^2+a+1 , 已知P(1,2) 為切線方程式的一點 , 則
令此方程式為 y - b = m (x- a) ,
y - (a^2+a+1) = (2a+1)(x-a)
y = 2ax - 2a^2 + x - a + a^2+a+1 = 2ax -a^2 + x -a +1 =
(2a+1)x + (1 - a^2  -a )
2 = 2a+1 + 1 - a^2  - a
2 = a + 2 -  a^2
0 = a -  a^2
0 = a(1-a)
a = 0 or a = 1
a = 0 , y = x +1
a= 1  , y = 3x -1 (標準答案有誤)

隱微分
2yy' = 1
y' = 1/2y , m = 1/4 , 所以切線方程式 y  - 2 = m (x-2) , y - 2 = 1/4 (x-2) ,
法線方程式 y - 2 = -1/m (x-2) , y - 2 = -4 ( x-2)

瞬間速度 = 極小時間變化量的位移比
加速度 =  極小時間變化量的速度比 , 即求速度的微分
v(t) = 4t^5 , a(t) = d/dt (v(t) ) = d/dt (4t^5) = 20t^4 , a(2) = 20 * 16 = 320

先求 y 的 x = 2 上的切線斜率, y'(2) = lim (4/x-1 - 4/2-1 ) / x -2  = 4 lim (1/x-1 - 1) / x -2 = 4 lim ( 1- x+1 ) / (x-2)(x-1)= 4 lim 2 - x / (2-x)(1-x)  = 4 lim 1/(1-x) = - 4 . 所以切線方程式 y - 4 = -4 (x-2)
y - 4 = -4x +8
y + 4x = 12

首先假設該切線方程式之切點為 (a,b) 其斜率為 m , m = 2x - 2 | x = a .
y - 1 = (2a-2) (x+1) , y - 1 = 2(a-1)(x+1)
y = 1 + 2(a-1)(x+1)