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I was watching and reading some online articles, and stubbled upon the power rule when dealing with derivatives. It states

$f(x) = x^n$, $n ≠ 0$
then
$f'(x) = nx^{n-1}$


I saw some proofs on this but no one addressed the reason why $n ≠ 0$. So I tested it out.


$f(x) = x^0 = 1$

thus, $f'(x) = 0 \times x^{0-1} = 0$


So I now saw why $n ≠ 0$, however knowing this, it sparked another question into my mind. What happens when $n$ is negative?


Once again I tested this out,
$f(x) = x^{-1}$
$f'(x) = -1 \times x^{-1 - 1}$


if you look at the last line you, a question appears, if $x = 0$ then,
$f'(x) = -1 \times \frac{1}{0^2}$


Dividing by zero is undefined, thus shouldnt the power rule be not defined for $x = 0$ for values $n < 0$. (This problem also applies for the case where $n = 0$)


Can anyone give a explanation for all this stuff? Thx!

Gary
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Eddie Wang
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1 Answers1

1

Yes, the rule works for negative powers too. (And as you say, the derivative might not be defined at $x=0$. That's ok! Nobody says that the derivative of a function must exist at all points, and of course, $x^{-n}$ has a vertical asymptote at $x=0$, so it makes sense that derivatives don't exist there.)

You can prove the power rule for negative exponents using the definition of the derivative. Or, if you're convinced that the power rule works for positive exponents, you can extend it to negative exponents using the product rule: for $n > 0$,

$$0 = \frac{d}{dx} 1 = \frac{d}{dx}\left( x^n x^{-n}\right)$$ so $$0 = nx^{n-1} x^{-n} + x^n \frac{d}{dx}\left(x^{-n}\right)$$ or $$-nx^{-1-n} = \frac{d}{dx}\left(x^{-n}\right).$$

Perhaps it's better to write the LHS as $$\frac{d}{dx}\left(x^{-n}\right) = (-n)x^{(-n)-1}$$ and now you can clearly see the power rule takes exactly the same form as for positive exponents.

user7530
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