# How to calculate a derivative using the “Power Rule” If it includes a negative exponent?

So my understanding of the power rule is that you take your problem with an exponent like this:

$x^5 = 5x^4$

or for $x^n$, $f'(x)=nx^{n-1}$

However, it does not seem to be working for me when applied to a problem with a negative exponent.

$q^{-3} -> -3q^{-2}$ is not a correct answer. I have tried converting it to a fraction also:

$q^{-3} -> \frac{1}{q^3}$ -> power rule: $\frac{1}{2q^2}$ is not correct.

How can we find the derivative of a power function when the exponent is negative?

Thanks

• $(-3) - 1 = -4 \neq -2$ – Daniel Fischer Jun 26 '14 at 22:37
• In the same manner. $f(x) = x^{-n} \implies f'(x) = -nx^{-n-1}$. – Kaj Hansen Jun 26 '14 at 22:38

You're trying to mix two approaches together, and as a result you're not getting the correct answer.

You can apply the power rule to $q^{-3}$ and find that its derivative with respect to $q$ is $$-3q^{-3-1}=-3q^{-4}.$$ Alternately, you can rewrite it as $\frac1{q^3}$ and apply the quotient rule, to see that its derivative is $$\frac{q^3\cdot0-1\cdot3q^2}{(q^3)^2}=\frac{-3q^2}{q^6}=-3q^{2-6}=-3q^{-4}.$$

The upshot is that if you want to use the power rule, you need to keep it in the appropriate form.

• @Cyllindra: Thanks for catching that! – Cameron Buie Jun 26 '14 at 22:52

$$(q^{-3})'=-3q^{-3-1}=-3q^{-4}=\frac{-3}{q^4}$$ $$(x^n)'=nx^{n-1},n\in\mathbb R$$

The power rule, i.e. the fact that $$\frac{d}{dx}[x^\alpha]=\alpha x^{\alpha-1},$$ is true for all $\alpha \in \mathbb{R} \setminus \{ 0\}$.

I normally use $\alpha$ instead of $n$ in this case, as $n$, to me, is an integer, whereas the power rule is true for all (nonzero) real $\alpha$.