∵f’(x)
=(x-100)’x(x-1)…(x-99)
+(x-100)[x(x-1)…(x-99)]’
而当x=100时,不管中括号的导数的值怎样,上式的第二部分积为0
∴f’(100)=100×99×…×1
=100!
f(x)=x(x-1)(x-2)…(x-99)(x-100),求f'(100)
3个回答
相关问题
-
已知f(x)=(x-1)(x-2)…（x-100)求f'(99)
-
f(x)=x(x_1)(x_2)…(x_99)(x_100) 求F（0）导数
-
f(x)=(x-1)(x-2)(x-3).(x-99)(x-100)求f(1)的导数
-
设函数f(x)=x*(x-1)*(x-2)...(x-99)(x-100),则f′(0)=
-
f(x)=x(x-1)(x+2)(x-3)(x+4)……(x+100),求f'(1)
-
f(x)=2x/1+2x 求f(1)+f(2).+f(100)+f(1/2)+f(2/2)+...+f(100/2)+.
-
当X=-1时 求X+2X^2+3X^3+…+99^99+100X^100
-
已知：f（x）=(x-1)(x-2)(x-3)……（x-100)求f'(1)
-
当x=-1时,求x+2x²+3x³+……+99x^99+100x^100的值
-
100X100—99X99+98X98+...+2X2—1X1