Page:Calculus Made Easy.pdf/43

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

$dy=5x^{4}dx$,

whence

${\frac {dy}{dx}}=5x^{4}$,

exactly as we supposed.

 

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

Let

$y=x^{n}$,

then, we should expect to find that

${\frac {dy}{dx}}=nx^{(n-1)}$.

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

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

Case of a negative power.

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

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