一道关于导数的数学题

来源：百度知道 编辑：UC知道 时间：2024/06/08 17:20:57
f(x)=(x+1)(x+2)(x+3).......(x+10),求f'(1).
帮忙解决一下这个问题,跪谢~

f'(x)=f(x)*[1/(x+1)+2/(x+2)+3/(x+3)+...+(10/(x+10)]

f'(1)=11!*(1/2+1/3+1/4+...+1/11)
=11!*(1/2+1/3+...+1/11)

f(x)=(x+1)(x+2)(x+3).......(x+10)的导数是
f'(x)=(x+1)'(x+2)(x+3).......(x+10)+(x+1)(x+2)'(x+3).......(x+10)+(x+1)(x+2)(x+3)'.......(x+10)+……+(x+1)(x+2)(x+3).......(x+10)'
=(x+2)(x+3).......(x+10)+(x+1)(x+3).......(x+10)+(x+1)(x+2)(x+4).......(x+10)+……+(x+1)(x+2)(x+3).......(x+9)
所以
f'(1)=(1+2)(1+3).......(1+10)+(1+1)(1+3).......(1+10)+(1+1)(1+2)(1+4).......(1+10)+……+(1+1)(1+2)(1+3).......(1+9)
=80627040

两边同取自然对数:lnf(x)=ln[(x+1)…(x+10)]对两边求导:f'(x)/f(x)=1/(x+1)+1/(x+2)+…+1/(x+10)所以f'(1)=(2*3*4*…*11)*(1/2+1/3+…+1/11)