year 400==0||(year 4==0 &&年份 year != 0
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/07 06:30:01
如果你能把表格描述一下或者发上来,就更好给你解释了.现在来看,就是如果“D955=0”,结果就是0.如果“D955不等于0”,就比较D955和YEAR($B$3),YEAR($B$3)小于0时结果等于
关于y求偏导,得f'y=e^x+1.则则f`y(0.0)=2再问:详细点的计算再答:在求y偏导的过程中,凡是与x有关的代数式均视为常熟,意思是ye^x中e^x为常熟,所以这部分关于y求偏导,为e^x,
Ye-ye-ye-ye-yeahohhhhyeahohhhYe-ye-ye-ye-yeahohhhhohhhYe-ye-ye-ye-yeahohhhhyeahohhhYe-ye-ye-ye-yeaho
fx(x,y)=yx^(y-1)e^x+x^ye^x;fx(1,-x)=-xe+e=e(1-x)
前一个题目两边同时求导,也太简单了.第二个设y=x^5+x-1dy=5x^4+1,全域恒正,所以Y单调递增(R上的单调函数),由于X=0时Y=-1,x=1时y>0,所以,根据连续函数零值定理,在X=0
xe^y+ye^x=0直接对x求导x'*e^y+x*(e^y)'+y'*e^x+y*(e^x)'=0e^y+x*e^y*y'+y'*e^x+y*e^x=0e^y+(xe^y+e^x)*y'+ye^x=
再问:。再答:怎么了?
两边对x求导,把y看成是x的复合函数:lny+xy'/y+y'e^(xy)+ye^(xy)(y+xy')=0y'[x/y+e^(xy)+xye^(xy)]=-lny-y²e^(xy)得y'=
EX=∫[0,+∞]xe^(-x)dx∫[0,+∞]ye^(-y)dy=1.E(X^2)=∫[0,+∞]x^2e^(-x)dx∫[0,+∞]ye^(-y)dy=2.EY=∫[0,+∞]e^(-x)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^(
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
1.hasbeendeadis,since2.hasbeen,for3.hasbeenaway,haspast4.havehad,for5.hasbeenin6.Shesaidshehadworked
再问:采用复合函数求导法,怎么求再答:
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)
是yeah,欢呼的意思.有时也是yes的口语化,表示赞同,ok的意思
两边求微分的2xdx+2zdz=2e^zdy+2ye^zdz解得dz=(2e^zdy-2xdx)/(2z-2ye^z)=(e^zdy-xdx)/(z-ye^z)
z=x²ye^y那么∂z/∂x=2xye^y∂z/∂y=x²e^y+x²ye^y所以二阶偏导数为∂²
x,y随t增减趋势,大致画出图像是从A(1,0) 沿着逆时针到B(1,-2π)的一段曲线..设原题目中P=y+ye^x,Q=x+e^x因为Q'x=P'y,所以原积分与路径无关