and subtracting ${\displaystyle y=x^{5}}$ leaves us

whence

exactly as we supposed.

Following out logically our observation, we should conclude that if we want to deal with any higher power,—call it ${\displaystyle n}$—we could tackle it in the same way.

Let

then, we should expect to find that

For example, let ${\displaystyle n=8}$, then ${\displaystyle y=x^{8}}$; and differentiating it would give ${\displaystyle {\frac {dy}{dx}}=8x^{7}}$.

And, indeed, the rule that differentiating ${\displaystyle x^{n}}$ gives as the result ${\displaystyle nx^{n-1}}$ is true for all cases where ${\displaystyle n}$ is a whole number and positive. [Expanding ${\displaystyle (x+dx)^{n}}$ by the binomial theorem will at once show this.] But the question whether it is true for cases where ${\displaystyle n}$ has negative or fractional values requires further consideration.

Case of a negative power.

Let ${\displaystyle y=x^{-2}}$. Then proceed as before:

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