已知函数f(x)=(x+2)ex次方,则f1(0)=
问题描述:
已知函数f(x)=(x+2)ex次方,则f1(0)=
答
f'(x)=(x+2)'*e^x+(x+2)*(e^x)'=e^x+(x+2)e^x
f'(0)=e^0+(0+2)*e^0=1+2=3