已知函数y=x^2ln(x^2)+cosx,求它的微积分dy

2025-02-25 00:58:42
推荐回答(1个)
回答1:

dy=ln(x^2)·d(x^2)+x^2·dln(x^2)-sinxdx

=2x·ln(x^2)dx+x^2·1/x^2·d(x^2)-sinxdx

=2x·ln(x^2)dx+2xdx-sinxdx

=[2x·ln(x^2)+2x-sinx]dx