若d f(1/x^2) /dx =1/x,求 f'(x)=?

d f(1/x^2) /dx =1/x
f'(1/x^2)*(-2/x^3）=1/x
f'(1/x^2)=-x^2/2
令1/x^2=t
f'(t)=-1/2t
f'(x)=-1/2x

令1/x=k
则df(k^2)/d(1/k)=k
df(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)