设u=√x则f（√x）=cos2x 化为 f（u）=cos2u^2有微分公式：df/dx=df/du*du/dx则f'(x)=df/du*du/dx=(cos2u^2)'*(√x)'=-sin2u^2*(2u^2)'*[1/(2√x)]=-4u*sin2u^2*[1/(2√x)]将u=√x代回f'(x)=-4√xsin2x/2√x=-2sin2x...