求導法一條!唔識bo.....

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圖片參考：http://img405.imageshack.us/img405/241/86438591.png


2010-11-29 21:24:03 補充：
http://img405.imageshack.us/img405/241/86438591.png

2010-11-29 21:27:36 補充：
Correction : h(x)=ln(x^2x + 1)
h'(x) 1/(x^2x + 1)[2(ln x+1)x^2x]
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http://img528.imageshack.us/img528/3724/42471757.png

2010-11-30 22:33:30 補充：
對,你一定要設y=x^2x才可以求導

Sol
h'(x)=dh(x)/dx
=[dln(x^(2x)+1)/d(x^(2x)+1)]*[d(x^(2x)+1)/dx^(2x)]*[dx^(2x)/dx]
=1/[(x^(2x)+1]*[dx^(2x)/dx]
Set y=x^(2x)
lny=2xlnx
dx^(2x)/dx
=[dx^(2x)/dln(x^(2x))]*[dln(x^(2x)/dx]
=[dy/dlny]*[d(2xlnx)/dx]
=2[dlny/dy]^(－1)*[d(xlnx)/dx]

2010-11-29 21:40:25 補充：
=2(1/y)^(－1)*[xdlnx/dx+lnxdx/dx]
=2y*[x*(1/x)+lnx]
=2x^(2x)*(1+lnx)
So
h'(x)=1/[(x^(2x)+1]*[dx^(2x)/dx]
=1/[(x^(2x)+1]*2x^(2x)*(1+lnx)
=2x^(2x)*(1+lnx)/[(x^(2x)+1]
h'(2)=2*2^(4)*(1+ln2)/[(2^(4)+1]
=(32/17)*(1+ln2)