一道关于导数的选择题若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数,则f'(1)=( )A.24 B.-24 C.10 D.-10
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![一道关于导数的选择题若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数,则f'(1)=( )A.24 B.-24 C.10 D.-10](/uploads/image/z/5573595-3-5.jpg?t=%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E5%AF%BC%E6%95%B0%E7%9A%84%E9%80%89%E6%8B%A9%E9%A2%98%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3D%28x-1%29%28x-2%29%28x-3%29%28x-4%29%28x-5%29%2C%E4%B8%94f%27%28x%29%E6%98%AF%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%AF%BC%E5%87%BD%E6%95%B0%2C%E5%88%99f%27%281%29%3D%28+%29A.24+B.-24+C.10+D.-10)
一道关于导数的选择题若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数,则f'(1)=( )A.24 B.-24 C.10 D.-10
一道关于导数的选择题
若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数,则f'(1)=( )
A.24 B.-24 C.10 D.-10
一道关于导数的选择题若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数,则f'(1)=( )A.24 B.-24 C.10 D.-10
f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)
f'(x)=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-1)(x-2)(x-4)(x-5)+ (x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-3)(x-4)
f'(1)=(1-2)(1-3)(1-4)(1-5)+0+0+0+0
=24
所以选A
f(1)=0
f'(1)=[f(1+deta)-f(1]/deta=f(1+deta)/deta
代入f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),得
f'(1)=deta*(-1+deta)*(-2+deta)*(-3+deta)*(-4+deta)/deta=(-1+deta)*(-2+deta)*(-3+deta)*(-4+deta)
deta约等于0
所以结果是24
A.因为如果带1进去就是0,所以剩f'(x)=(-1)*(-2)*(-3)*(-4)=24
g(x)=(x-2)(x-3)(x-4)(x-5)
f(x)=(x-1)g(x)
f'(x)=g(x)+(x-1)g'(x)
f'(1)=g(1)=4!*(-1)^4=24
选A