求微分f(x)=x(x+1)(x+2)……(x+n),则f'(0)=?设x+y=tany,则dy=
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/02 18:22:40
求微分f(x)=x(x+1)(x+2)……(x+n),则f'(0)=?设x+y=tany,则dy=
求微分f(x)=x(x+1)(x+2)……(x+n),则f'(0)=?
设x+y=tany,则dy=?
求微分f(x)=x(x+1)(x+2)……(x+n),则f'(0)=?
设x+y=tany,则dy=?
f'(x)
={x[(x+1)(x+2)……(x+n)]}'
=(x)'[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
=[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
令P(x)=[(x+1)(x+2)……(x+n)]'
f'(0)=1*2*..*n+0*P(0)=n!
x+y=tany
两边同时求微分dx+dy=dy/cos(y^2)
解得dy=1/tan(y^2)
={x[(x+1)(x+2)……(x+n)]}'
=(x)'[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
=[(x+1)(x+2)……(x+n)]+x[(x+1)(x+2)……(x+n)]'
令P(x)=[(x+1)(x+2)……(x+n)]'
f'(0)=1*2*..*n+0*P(0)=n!
x+y=tany
两边同时求微分dx+dy=dy/cos(y^2)
解得dy=1/tan(y^2)
tany=x+y 求dy/dx
设f(x)=1/x,y=f[(x-1)/(x+1)],求dy/dx
求dy/dx-y/x=tany/x的通解
y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy
设函数y=f(x)有f'(x.),则当Δˇx→0f(x)在x=xˇo处的微分dy是
设y=f^2(3x-2/3x+2),f(x)=In(1+x^2),则dy/dx|x=0=?
设y=f(-x^2) 则dy=
设f(x)可导,且f'(0=1,又y=f(x^2+sin^2x)+f(arctanx),求dy/dx /x=0
设y=(2+x)^x,(x>0) ,求dy
设y=f(sinx)+e^x^2,f'(x)存在,求y'及dy
设F(x)可导,y=f(x^2),则dy/dx=?
解dy/dx=y/x+tany/x 的微分方程