链式法则[f(g(x))]'=f'(g(x))g'(x)此处f(x)=ln xg(x)=cosx+sinx所以代入可得f'(x)=1/xf'(g(x))=1/g(x)=1/(cosx+sinx)g'(x)=-sinx+cosx所以[ln(cosx+sinx)]'=[1/(cosx+sinx)]*[-sinx+cosx]=(-sinx+cosx)/(cosx+sinx)...