.设1=Sdx[_f(x,y)dy，写出改变积分次序后的二次积分

设y=（x/1-x)^x,求dy/dx设y=（x/1-x)^x,求dy/dx设y=（x/1-x)^x,求dy/dx先两边取ln,得到lny=xln(x/1-x),然后两边求导,（dy/dx)*(1/y)=ln(x/1-x)+1/1-x.最后

设y=ln(x/1+x)-cot2x,求dy.设y=ln(x/1+x)-cot2x,求dy.设y=ln(x/1+x)-cot2x,求dy.y=ln(x/(1+x))-cot2xdy=[(1+x)/x]d(x/(1+x))+(csc2x)^2

设y=(1/x)^x,求dy设y=(1/x)^x,求dy设y=(1/x)^x,求dylny=xln(1/x)对x求导(1/y)y'=ln(1/x)+x*x*(-1/x²)y'=y*[ln(1/x)-1]y'=(1/x)^x*[ln

设y=e^(-x)cos1/x,求dy.设y=e^(-x)cos1/x,求dy.设y=e^(-x)cos1/x,求dy.

设y=ln(1+x^2)则dy=设y=ln(1+x^2)则dy=设y=ln(1+x^2)则dy=

设y=e^(-1/x),则dy=?设y=e^(-1/x),则dy=?设y=e^(-1/x),则dy=?dy=（e^(-1/x)）*（-1/x）dx=（e^(-1/x)）*（1/x*x）dx

设y=arctan1/x,则dy=rt设y=arctan1/x,则dy=rt设y=arctan1/x,则dy=rt-1/(x^2+1)

设y=e(-1/x)次方,则dy=设y=e(-1/x)次方,则dy=设y=e(-1/x)次方,则dy=dy=(x平方之一)·e(-1/x)dx

设y=In(1+2x)则dy=设y=In(1+2x)则dy=设y=In(1+2x)则dy=dy=dln(1+2x)=[1/(1+2x)]d(1+2x)=[2/(1+2x)]dx

设函数y=cos2x/(x-1)求dy设函数y=cos2x/(x-1)求dy设函数y=cos2x/(x-1)求dydy=d[cos2x/(x-1)]=y'dx={[-2sin2x*(x-1)-cos2x]/(x-1)^2}dx=-[2sin

设y=√(x^2-1)求dy/dx设y=√(x^2-1)求dy/dx设y=√(x^2-1)求dy/dx这是最简单的复合函数y=√uu=x²-1dy/dx=dy/du*du/dx=(1/2√u)*2x=x/2√x²-1

设y-x-1/2sinx,则dx/dy=设y-x-1/2sinx,则dx/dy=设y-x-1/2sinx,则dx/dy=是y=x-1/2sinx吧那么y'=1-(1/2)cosx又y'=dy/dx=1-(1/2)cosx所以dx/dy=1/

设y=sin2x/(1+x^2),求dy/dx,设y=sin2x/(1+x^2),求dy/dx,设y=sin2x/(1+x^2),求dy/dx,dy/dx=[(sin2x)'(1+x²)-sin2x·(1+x²)']/(

设y=cos^2(x+1),求dy设y=cos^2(x+1),求dy设y=cos^2(x+1),求dy楼上的少写了“-”和“dx”吧dy=2cos(x+1)•[-sin(x+1)]dx=-sin2(x+1)dx

设y=ln(1+e^-x)求dy设y=ln(1+e^-x)求dy设y=ln(1+e^-x)求dy对等式两边同时求导:dy/dx=-e^-x/(1+e^-x)dy=-1/(1+e^+x)

设y=(1-x)³cosx,求dy/dx设y=(1-x)³cosx,求dy/dx设y=(1-x)³cosx,求dy/dx简单的一次求导问题

设y=e^1-2xsinx,求dy设y=e^1-2xsinx,求dy设y=e^1-2xsinx,求dy你问的是y=(e^1-2x)sinx吧?y'=-2(e^1-2x)sinx+(e^1-2x)cosxy'=-(2sinx+2xcosx)*

设y=sin2x/(1+x^2),求dy/dx,设y=sin2x/(1+x^2),求dy/dx,设y=sin2x/(1+x^2),求dy/dx,

设y=e∧1/x+sin2x,求dy设y=e∧1/x+sin2x,求dy设y=e∧1/x+sin2x,求dy点击看大图

设y=e^cos^2*1/x,求dy设y=e^cos^2*1/x,求dy设y=e^cos^2*1/x,求dylny=cos^2(1/x)两边求导:1/ydy/dx=2cos(1/x)[cos(1/x)]'1/ydy/dx=2cos(1/x)