1.利用對數求導法求導:y=[(x-1)x(x+1)(x+2)/(x-2)^2(x+3)]^1/3 2.求二階導數（y`

題目：

1.利用對數求導法求導:y=[(x-1)x(x+1)(x+2)/(x-2)^2(x+3)]^1/3 2.求二階導數（y``）：y=(x^2-2x+5)^10
1.求極限：lim（x趨向0+）lnx/lnsinx
2.單調區間：y=x-e^x

解答：

1.lny=1/3[ln(x-1)+lnx+ln(x+1)+ln(x+2)-2ln(x-2)-ln(x+3)]
y'/y=1/3[1/(x-1)+1/x+1/(x+1)+1/(x+2)-2/(x-2)-1/(x+3)]
y'=1/3[1/(x-1)+1/x+1/(x+1)+1/(x+2)-2/(x-2)-1/(x+3)][(x-1)x(x+1)(x+2)/(x-2)^2(x+3)]^1/3
2.y'=10(x^2-2x+5)^9(2x-2)=20(x^2-2x+5)^9(x-1)
y"=20[(x^2-2x+5)^9]'(x-1)+20(x^2-2x+5)^9[(x-1)]'
=20[9(x^2-2x+5)^8(x-1)²]+20(x^2-2x+5)^9
=20(x²-2x+5)^8[9(x-1)²+x²-2x+5]
3.lim（x趨向0+）lnx/lnsinx =lim 1/x /(cosx/sinx)=lim sinx/(xcosx)=lim tanx/x=1
4.y'=1-e^x>=0 e^x