ye^x lny=1-搜答案-智慧作业帮

最后乘以dy/dx实际上是对Iny中的y求导,因为Iny是复合函数（y是关于x的函数）,所以(Iny）'=1/y*y'=1/y*dy/dx

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fx(x,y)=yx^(y-1)e^x+x^ye^x;fx(1,-x)=-xe+e=e(1-x)

答：y=xlny+x^3对x求导：y'=lny+(x/y)y'+3x^2(1-x/y)y'=lny+3x^2y'=(3x^2+lny)y/(y-x)所以：dy/dx=(3x^2+lny)y/(y-x)

是汀汀乐队的么?

直接对方程关于x的函数求导：2x+2yy'-lny-xy'/y=0整理一下：y'=dy/dx=(2x-lny)/(x/y-2y)dy=(2x-lny)/(x/y-2y)dx再问：求导那里应该是2x-2

再问：。再答：怎么了？

xe^2y-ye^2x=1,求dy/dx

∵xe^(2y)-ye^(2x)=1==>e^(2y)dx+2xe^(2y)dy-e^(2x)dy-2ye^(2x)dx=0(等式两端取微分)==>[2xe^(2y)-e^(2x)]dy=[2ye^(

x(lny(x))'+lny(x)+y(x)(e^(xy(x)))'+y'(x)e^(xy(x))=0x(1/y(x))y'(x)+lny(x)+y(x)(e^(xy(x)))(xy(x))'+y'(

两边同时对x求导利用积法则+复合求导(dy/dx)e^x+ye^x+(1/y)*dy/dx=0(dy/dx)(e^x+1/y)=-ye^xdy/dx=-ye^x/(e^x+1/y)ye^x=1-lny

两边x求导得y'e^x+ye^x+y'/y=0y'=-ye^x/(e^x+1/y)=-y^2e^x/(ye^x+1)y''=[(-2yy'e^x-y^2e^x)(ye^x+1)+y^2e^x(y'e^

再问：采用复合函数求导法，怎么求再答：

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ye^x-xe^y=0求导dy/dx

其实就是隐函数求导,方程两边同时对x求导,y看做中间变量y'e^x+ye^x-e^y-(xe^y)y'=0所以dy/dx=y'=(e^y-ye^x)/(e^x-xe^y)

ye^x*log(ye)

ye,ye是什么意思?

是yeah,欢呼的意思.有时也是yes的口语化,表示赞同,ok的意思

原式=∫[1,2]dx∫[1/x,2]ye^(xy)dy=∫[1,2]dx∫[1/x,2]y/xe^(xy)d(xy)第一个对y的积分中x是常数=∫[1,2]1/xdx∫[1/x,2]yde^(xy)

P在y=-x+6上P是第一象限所以-x+6>0,x>00<x<6三角形底边是AO（1）第一象限的P点纵坐标y即为△PAO边OA对应的高S=|OA|*y/2=5(-