Question

f(x)=x^2e^ax的导数,

Answer 1

用乘法则(vu)'=uv'+vu'和链式法则:[v(u(x))]'=v'(u(x))*u'(x)

f(x)=x²*e^(ax)

f'(x)=e^(ax)*(x²)'+x²*(e^(ax))'

=e^(ax)*2x+x²*e^(ax)*(ax)'

=2xe^(ax)+ax²e^(ax)

=(2+ax)xe^(ax)