设函数f(x)=x(x-1)(x-2)……(x-n)，则f'(1)=?

设f(x)=x(x+1)(x+2)……(x+n)则f'( - 1)= - (n - 1)!请写出具体过程

f'(x)=[x(x+1)(x+2)……(x+n)]'=[x+1][x*(x+2)……(x+n)]=[x*(x+2)……(x+n)]+[x+1]*[x*(x+2)……(x+n)]f'(-1)=x*(x+2)……(x+n)=-(n-1)!

设f(x)=x(x - 1)(x - 2).(x - n)求fx的n+1阶导函数)

f(x)的最高次项是x^(n+1),其余项的次数均低于(n+1),n+1阶求导后均为0 x^(n+1)一阶导数=(n+1)x^n x^(n+1)二阶导数=(n+1)nx^(n-1).x^(n+1)n+1阶导数=(n+1)!x^(1-1)=(n+1)!∴f(x)=x(x-1)(x-2).(x-n)的n+1阶导函数=(n+1)!

设函数f(x - 1)=x^2 - x,则f(x)=()

很简单 解:(1)对于函数f(x)=ax-(a+1)ln(x+1) 其一次导函数为f'(x)=(ax-1)/(x+1) 二次导函数为f''(x)=(a+1)/(x+1)² 易知 当a>-1时,f'(x)单调递增,所以只需要f'(x)≥f'(2.

设f(x)=x(x+1)(x+2)...(x+n),求f '( - 1).

f(x)=x(x+1)(x+2)……(x+n) f'(x)=(x+1)(x+2)……(x+n)+x[(x+2)(x+3).(x+n)+(x+1)(x+3)..(x+n)+.] f'(0)=n!+0=n!

设f(x)=|x - 1| - |x|,则f(1/2)等于

f(1/2)=|-1/2|-|1/2|=1/2-1/2=0所以原式=f(0)=|-1|-|0|=1

设f(x)=|x - 1| - |x|,则f[f(1/2)]等于

f(2)=√(2-1)=1f(1)=√(1-1)=0f(0)=1f{ f [ f(2) ] }=f[f (1)] =f(0)=1

设函数f(x)=|x - 1| - 2, |x|≤1;1+x分之1, |x|>1,则f(f(f(2分之1)).

f(x)=|x-1|-2 (|x|<=1,-1=<x<=1) f(x)=1/(1+x) (|x|>1,x<-1或者x>1) f(1/2)=|1/2-1|-2=1/2-2=-3/2 f(f(1/2))=f(-3/2)=1/(1+(-3/2))=-2 f(f(f(1/2)))=f(-2)=1/(1+(-2))=-1

设函数f(x)=x(x - 1)(x - 2)(x - 3)(x - 4),则f'(0)

选a 把(x-1)(x-2)(x-3)看也一个,高g(x); 则f(x)=(x-4)*g(x); 所以f'(x)=g(x)+g'(x)*(x-4); 则有f'(4)=g(4)>0; 同样的方法,可得: f'(1)0;(令g(x)=(x-1)(x-3)(x-4)) f'(3)

设fx=x(x - 1)(x - 2).(x - n)求导

f(x)=(x-1)(x-2)……(x-n)/(x+1)(x+2)……(x+n) =(x-1)[(x-2)……(x-n)/(x+1)(x+2)……(x+n)] =(x-1)'[(x-2)……(x-n)/(x+1)(x+2)……(x+n)]+(x-1)[(x-2)……(x-n)/(x+1)(x+2)……(x+n)]'f(1)=(1-2)……(1-n)/(1+1)(1+2)……(1+n) =(-1)^(n-1)*1*2*.*(n-1)/2*3*.*(n+1)=(-1)^(n-1)/(n(n+1)),

高等数学f(x)=x(x+1)(x+2)…(x+n),n属于N,求f'(0)=.

解由f(x)=x(x+1)(x+2)…(x+n) 知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)]' 故f'(0)=[(0+1)(0+2)…(0+n)]+0*[(0+1)(0+2)…(0+n)]'=1*2*3..*n=n!.