Question

(ax+b)^x形式的函数导数怎么求?a、b为常数

Answer 1

y=(ax+b)^x

y=e^[ln(ax+b)*x]

y的导数={e^[ln(ax+b)*x]}*[x/(ax+b)+ln(ax+b)]

==(ax+b)^x][x/(ax+b)+ln(ax+b)]