這四題微分怎麼算?

2011-07-13 9:16 am
Find dx/dy

1.y = [(x+1)/(x-3)]^4

2.x^2 - 2xy^2 = y^4

3.xcosy + ycosx = 1

4.y = tan(sinx^2)

回答 (1)

2011-07-13 11:28 am
✔ 最佳答案
1.
dy/dx
= (d/dx)[(x+1)/(x-3)]^4
= (d/dx)[(x+1)^4/(x-3)^4]
= [(x-3)^4•d(x+1)^4/dx - (x+1)^4•d(x-3)^4/dx] / (x-3)^8
= [(x-3)^4•4(x+1)^3 - (x+1)^4•4(x-3)^3] / (x-3)^8
= 4(x-3)^3•(x+1)^3[(x-3)-(x+1)] /(x-3)^8
= -16(x+1)^3 / (x-3)^5


2.
x^2 - 2xy^2 = y^4
(d/dx)(x^2 - 2xy^2) = (d/dx)y^4
2x - [2x•2y•(dy/dx) + 2y^2•(dx/dx)] = 4y^3•(dy/dx)
2x - 4xy•(dy/dx) - 2y^2 = 4y^3•(dy/dx)
(4xy + 4y^3)•(dy/dx) = 2x - 2y^2
4y(x + y^2)•(dy/dx) = 2(x - y^2)
dy/dx = (x - y^2) / 2y(x + y^2)


3.
xcosy + ycosx = 1
(d/dx)(xcosy + ycosx) = (d/dx)(1)
x•dcosy/dx + cosy•dx/dx + y•dcosx/dx + cosx•dy/dx = 0
x(-siny)•(dy/dx) + cosy + y(-sinx) +cosx•(dy/dx) = 0
(xsiny - cosx)•(dy/dx) = cosy - ysinx
dy/dx = (cosy - ysinx) / (xsiny - cosx)


4.
dy/dx
= dtan(sinx^2)/dx
= dtan(sinx^2)/dsinx^2 • dsinx^2/dx^2 • dx^2/dx
= sec^2(sinx^2) • cosx^2 •2x
= 2x cosx^2 sec^2(sinx^2)
參考: 賣女孩的火柴


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