Expanding this by the binomial theorem (see p. 141), we get

${\displaystyle =x^{-2}\left[1-{\frac {2dx}{x}}+{\frac {2(2+1)}{1\times 2}}{\left({\frac {dx}{x}}\right)}^{2}-etc.\right]}$

${\displaystyle =x^{-2}-2x^{-3}\cdot dx+3x^{-4}(dx)^{2}-4x^{-5}(dx)^{3}+etc}$.

So, neglecting the small quantities of higher orders of smallness, we have:

${\displaystyle y+dy=x^{-2}-2x^{-3}\cdot dx}$.

Subtracting the original ${\displaystyle y=x^{-2}}$, we find

${\displaystyle dy\,\,=-2x^{-3}dx}$,

${\displaystyle {\frac {dy}{dx}}=-2x^{-3}}$.

And this is still in accordance with the rule inferred above.

Case of a fractional power.

Let ${\displaystyle y=x^{\tfrac {1}{2}}}$. Then, as before,

${\displaystyle y+dy=(x+dx)^{\tfrac {1}{2}}=x^{\tfrac {1}{2}}\left(1+{\frac {dx}{x}}\right)^{\tfrac {1}{2}}}$

${\displaystyle ={\sqrt {x}}+{\frac {1}{2}}{\frac {dx}{\sqrt {x}}}-{\frac {1}{8}}{\frac {(dx)^{2}}{x{\sqrt {x}}}}+}$ terms with higher powers of ${\displaystyle dx}$.

Subtracting the original ${\displaystyle y=x^{\tfrac {1}{2}}}$, and neglecting higher powers we have left:

⁠${\displaystyle dy={\frac {1}{2}}{\frac {dx}{\sqrt {x}}}={\frac {1}{2}}x^{-{\tfrac {1}{2}}}\cdot dx}$,

and ⁠${\displaystyle {\frac {dy}{dx}}={\frac {1}{2}}x^{-{\tfrac {1}{2}}}}$. Agreeing with the general rule.