differentiation second derivative

Question

第3题 之前忘写了

Comments

其实你也已经问了很多次differentiation 了。你有自己尝试做吗？毕竟真正考试是不能问人的。

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我是真的不知道要问谁了才来这里问的
我也是没办法了

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我也不想一直这样问
我问的是网上找不到的
课本也没有比较详细的解释的

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不是不给你问，而是你自己应该也尝试先做。你可以放上你做的草稿，就算只是做了一半，还没解完的草稿也是可以。

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我也不是没想过怎么做
只是觉得这么做会麻烦到其他人
即麻烦别人要看我卡在哪里又要想怎么接我的接下去

Answer 1

1. Find (dy/dx)  (d²y/dx²)  [note: anx^n-1]

a) 

y = 3x⁴-5x²+2x-1

(dy/dx) = 3(4)x^4-1 - 5(2)x^2-1 +2(1)x^1-1 - 1(0)

(dy/dx) = 12x³-10x+2  [答案]

.

(dy/dx) = 12x³-10x+2

(d²y/dx²) = 12(3)x^3-1 -10(1)x^1-1 + 2(0)

(d²y/dx²) = 36x² - 10  [答案]

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b) 

y = 4x² - (2/x)

y = 4x² - 2x^-1

(dy/dx) = 4(2)x^2-1 - 2(-1)x^-1-1

(dy/dx) = 8x + 2x^-2

(dy/dx) = 8x + (2/x²)  [答案]

.

(dy/dx) = 8x + 2x^-2

(d²y/dx²) = 8(1)x^1-1 + 2(-2)x^-2-1

(d²y/dx²) = 8 - 4x^-3

(d²y/dx²) = 8 - (4/x³)  [答案]

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c)  [note: nab(bx+c)^n-1]

y = (3x+2)⁸

(dy/dx) = (8)(3x+2)^8-1(3)

(dy/dx) = 24(3x+2)⁷  [答案]

.

(dy/dx) = 24(3x+2)⁷

(d²y/dx²) = 24(7)(3x+2)^7-1(3)

(d²y/dx²) = 168(3x+2)⁶(3)

(d²y/dx²) = 504(3x+2)⁶  [答案]

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2) Find f'(x)  f''(x)

a)

f(x) = √x + (1/x²)

f(x) = x^1/2 + x^-2

f'(x) = (1/2)x^(1/2)-1 + (-2)x^-2-1

f'(x) = (1/2)x^-(1/2) - 2x^-3

f'(x) = 1/(2√x) - 2/x³  [答案]

.

f'(x) = (1/2)x^-(1/2) - 2x^-3

f''(x) = (1/2)(-1/2)x^-(1/2)-1 - 2(-3)x^-3-1

f'(x) = -(1/4)x^-(3/2) + 6x^-4

f'(x) = -1/(4√x²) + 6/x⁴  [答案]

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b)

f(x) = (x⁴+2)/x²

f(x) = (x⁴/x2) + (2/x²)

f(x) = x² + 2x^-2

f'(x) = (2)x^2-1 + 2(-2)x^-2-1

f'(x) = 2x¹ - 4x^-3

f'(x) = 2x - 4/x³  [答案]

.

f'(x) = 2x - 4x^-3

f''(x) = 2(1)x^1-1 - 4(-3)x^-3-1

f''(x) = 2 + 12x^-4

f''(x) = 2 + 12/x⁴  [答案]

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c)  {note: [(v)(du/dx) - (u)(dv/du)]/v²}

f(x) = (2x+5)/(x-1)

u = 2x+5       v = x-1

(du/dx) = 2   (dv/dx) = 1

.

f'(x) = [(v)(du/dx) - (u)(dv/du)]/v2

f'(x) = [(x-1)(2) - (2x+5)(1)]/(x-1)²

f'(x) = [(2x-2)-(2x+5)]/(x-1)²

f'(x) = (2x-2-2x-5)/(x-1)²

f'(x) = (-7)/(x-1)²  [答案]

.

f'(x) = (-7)/(x-1)²

f'(x) = (-7)(x-1)^-2

f''(x) = (-7)(-2)(x-1)^-2-1(1)

f''(x) = (14)(x-1)^-3

f''(x) = 14/(x-1)³  [答案]

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3) coordinate of A  /  value of A

y = x³+3x²-9x+2

(dy/dx) = (3)x^3-1 + (3)(2)x^2-1 - (9)(1)x^1-1 + 2(0)

(dy/dx) = 3x² + 6x - 9

.

Find coordinate x

(dy/dx) = 0

(dy/dx) = 3x² + 6x - 9

3x² + 6x - 9 = 0

(x-1)(x+3) = 0

x=1 , x=-3

.

Find coordinate y

y = x³+3x²-9x+2

[x=1]

y = (1)³ + 3(1)² - 9(1) + 2

y = -3

.

y = x³+3x²-9x+2

[x=-3]

y = (-3)³ + 3(-3)² - 9(-3) + 2

y = 29

.

[x=1 , y=-3] [x=-3 , y=29]

(1,-3) (-3,29)

.

(dy/dx) = 3x² + 6x - 9

(d²y/dx²) = 3(2)x^2-1 + 6(1)x^1-1 - 9(0)

(dy/dx) = 6x + 6

[x=1 , x=-3]

.

6(1)+6 = 12

6(-3)+6 = -12

.

12 , -12 

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https://youtu.be/399HZIKKzfU

课本里面的题目网上都会有解答方法

学习不可怕，只怕你没去学。好学不可怕，只怕你没去找。

以上都是我自己的算法 (没抄网上的解答方法)

Comments

知道了