数学题 帮忙解答 急！！

Question

请用chain rule 谢谢！form 5 chapter2 differentiation

Answer 1

petua rantai [chain rule] :

(dy/dx) = (dy/du) x (du/dx)

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y = (2x - [1/x])²

.

let u = 2x - (1/x)   ,  y = u²

(du/dx) = 2+(1/x²)  ,  (dy/du) = 2u

.

(note: anx^n-1 )

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

                =  2(1)x^1-1 - (-1)x^-1-1

                = 2 + x^-2

                = 2 + (1/x²)

.

u² = 1(2)u^2-1

     = 2u¹

     = 2u

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(dy/dx) = (dy/du) x (du/dx)

            = (2u) x (2+[1/x²])

            = 2(2x - [1/x]) x (2+[1/x²])

           = 2(2x²-1/x) x (2x²+1/x²)

           = [2(2x²-1) x (2x²+1)] / x³

           = [2(4x⁴-1)] / x³

           = 8x⁴-2 / x³    [答案]

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把2个挂号里面的变分数

有2个分数是乘所以分子乘分子，分母乘分母，第一个2乘上去

(a+b)(a-b) = a²-b²

(2x²-1)(2x²+1) = (2x²)² - 1²

                               = (4x⁴-1)

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以上是我自己的算法

Comments

还有一题可以帮我去作作吗 也是differentiation的