微積分的問題 求 y=x^(x^2 +1) 在(1，1)的切線方程式? [y=x的x平方+1次方]?
回答 (2)
2016-12-05 5:51 am
微積分的問題 求 y=x^(x^2 +1) 在(1，1)的切線方程式?
Sol
y=x^(x^2+1)
lny=(x^2+1)lnx
dx^(x^2+1)/dx
=｛dx^(x^2+1)/dln[x^(x^2+1)]｝* ｛dln[x^(x^2+1)]/dx｝
=(dy/dlny)* d[(x^2+1)lnx]/dx
=[1/(dlny/dy)]*[(x^2+1)dlnx/dx+lnx*d(x^2+1)/dx]
=[1/(1/y)]*[(x^2+1)*(1/x)+lnx*2x]
=y*[(x^2+2x^2lnx+1)/x]
=[x^(x^2+1)]*(x^2+2x^2lnx+1)/x]
y’(1)=1*(1+1)/1=2
y－1=2(x－1)
y－1=2x－2
2x－y+1=0
Sol
y=x^(x^2+1)
lny=(x^2+1)lnx
dx^(x^2+1)/dx
=｛dx^(x^2+1)/dln[x^(x^2+1)]｝* ｛dln[x^(x^2+1)]/dx｝
=(dy/dlny)* d[(x^2+1)lnx]/dx
=[1/(dlny/dy)]*[(x^2+1)dlnx/dx+lnx*d(x^2+1)/dx]
=[1/(1/y)]*[(x^2+1)*(1/x)+lnx*2x]
=y*[(x^2+2x^2lnx+1)/x]
=[x^(x^2+1)]*(x^2+2x^2lnx+1)/x]
y’(1)=1*(1+1)/1=2
y－1=2(x－1)
y－1=2x－2
2x－y+1=0
2016-12-01 1:35 am
y=x^(x^2 +1)
dy/dx=(x^2+1)(2x)x^(x^2)
dy/dx|x=1 =2(2)(1)
dy/dx|x=1 =4
y=4x+c
1=4+c
c=-3
y=4x-3
4x-y-3=0
dy/dx=(x^2+1)(2x)x^(x^2)
dy/dx|x=1 =2(2)(1)
dy/dx|x=1 =4
y=4x+c
1=4+c
c=-3
y=4x-3
4x-y-3=0
收錄日期: 2021-04-18 15:49:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161130035823AAm4Kw1