已知f(x)=(x-1)(x-2)(x-3)(x-4).(x-2011),求f'(2011)=?
f(x)=(x-1)(x-2)(x-3)(x-4).(x-2011)=[(x-1)(x-2)(x-3)(x-4).(x-2010)](x-2011)所以f'(x)=[(x-1)(x-2)(x-3)(x-4).(x-2010)]'(x-2011)+[(x-1)(x-2)(x-3)(x-4).(x-2010)](x-2011)'
已知f(x)=(x-1)(x-2)(x-3)(x-4)...(x-2011),求f'(2011)=?
=[(x-1)(x-2)(x-3)(x-4)...(x-2010)](x-2011)所以 f'(x)=[(x-1)(x-2)(x-3)(x-4)...(x-2010)]'(x-2011)+[(x-1)(x-2)(x-3)(x-4)...(x-2010)](x-2011)'=[(x-1)(x-2)(x-3)(x-4)...(x-2010)]'(x-2011)+[(x-1)(x-2)(x-3)(x-4)....
已知f(x)=(x-1)(x-2)…(x-2011),则f'(1)=?
f(x)=(x-1)(x-2)…(x-2011)则f'(x)=(x-2)…(x-2011)+(x-1)(x-3)…(x-2011)+(x-1)(x-2)(x-4)…(x-2011)+…+(x-1)(x-2)…(x-2010)则f'(1)=(1-2)…(1-2011)+0+0+…+0=(-1)(-2)…(-2010)=2010!
求一道数学导数题 设f(x)=(x-1)(x-2)...(x-2011).求f'(x)
f'(x)\/f(x) =∑(1≤k≤2011)[1\/(x-k)],所以,f'(x) =f(x)*∑(1≤k≤2011)[1\/(x-k)] = …….
X(X-1)(X-2)...(X-2011),f(2012)'x=2012!那么f(2011)'X=?
先把前面的看成一个整体，原式=[X（X-1）(X-2)...(X-2010)]*(X-2011),求导数时原式 =[X（X-1）(X-2)...(X-2010)]‘*(X-2011)+[X（X-1）(X-2)...(X-2010)]，把x=2011代入[X（X-1）(X-2)...(X-2010)]‘*(X-2011)=0，只有右边不为0，所以 f(2011)'X=20...
f(x)=x(x-1)(x-2)(x-3)…(x-2013),求f’(x)
记g(x)=(x-1)(x-2)..(x-2012)则f(x)=g(x)(x-2013)f'(x)=g'(x)(x-2013)+g(x)f'(2013)=g(2013)=2012*2011*...*2*1=2012!
若函数f(x)=(x-1)(x-2)(x-3)...(x-2013) 则f′(2013)=__
记g(x)=(x-1)(x-2)..(x-2012)则f(x)=g(x)(x-2013)f'(x)=g'(x)(x-2013)+g(x)f'(2013)=g(2013)=2012*2011*...*2*1=2012!
已知函数f(x)={当x <0时,f(x)=x-1;当x>=0时,f(x)=f(x-1),则f(2011)=
f(2011)=f(2011-1)=f（2010)...=f(-1)=(-1)-1=-2
f(x)=(x-1)(X-2)...(x-2012) 则f'(2012)= 求详解
f(x)=(x-1)(x-2)...(x-2012)f'(x)=(x-2)...(x-2012)+(x-1)(x-3)...(x-2012)+(x-1)(x-2)(x-4)..(x-2012)...+(x-1)...(x-2011)f'(2012)=0+0+...+2011*2010*...*3*2*1 =1*2*3*...*2011 =2011!
设函数f(x)=x(x-1)(x-2)...(x-2012),则f(2012)的导数
F'(X)=(x-1)(x-2)...(x-2012)+x(x-2)...(x-2012)+x(x-1)(x-3)...(x-2012)+……+x(x-1)……(x-2011)F'(2012)=0+0+0+……+2012*2011*2010*……*1 =2012！（2012的阶乘的意思）
