令1/x=k则df(k^2)/d(1/k)=kdf(k^2)=d(1/k)*k=-dk/k=-dk^2/(2k^2)把k^2换成x就得df(x)=-dx/(2x)f'(x)=-1/(2x)
d f(1/x^2) /dx =1/xf'(1/x^2)*(-2/x^3）=1/xf'(1/x^2)=-x^2/2令1/x^2=tf'(t)=-1/2tf'(x)=-1/2x
