dy/dx

2008-12-13 5:58 am
微分以下
where a and n are constants

1) y= ((1-x)^n) /(1+x)


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



thx

回答 (1)

2008-12-18 9:09 am
✔ 最佳答案
1)
y = (1 - x)n/(1 + x)
y = (1 - x)n(1 + x)-1

dy/dx
= (1 - x)n•d(1 + x)-1/dx + (1 + x)-1•d(1 - x)n/dx
= (1 - x)n•d(1 + x)-1/d(1 + x) + (1 + x)-1•d(1 - x)n/[-d(1 - x)]
= (1 - x)n[-(1 + x)-2] + (1 + x)-1[-n(1 - x)n-1]
= -(1 - x)n/(1 + x)2 - n(1 - x)n-1/(1 + x)
= -[(1 - x)n + n(1 + x)(1 - x)n-1]/(1 + x)2

=====
2)
y = xn(1 - x2)1/2

dy/dx
= d[xn(1 - x2)1/2]/dx
= xn•d(1 - x2)1/2/dx + (1 - x2)1/2•dxn/dx
= xn•d(1 - x2)1/2/d(1 - x2)•d(1 - x2)/dx + (1 - x2)1/2•nxn-1
= xn•(1/2)(1 - x2)-1/2•(-2x) + nxn-1(1 - x2)1/2
= -xn+1/(1 - x2)1/2 + nxn-1(1 - x2)1/2
= xn+1[n(1 - x)2 - 1]/(1 - x2)1/2
=


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