Differentiating X^N Where N Is A Negative Integer
Learn to differentiate reciprocal powers for Queensland Year 11 Mathematical Methods (QCAA). A term like one over x cubed can be rewritten with a negative index, so the ordinary power rule applies to it directly.
You will learn to rewrite fractions as negative indices, apply the power rule when the index is negative, split a quotient into separate terms, and find the gradient of a curve — extending differentiation beyond polynomials.
Every question with a fully worked solution.
- Differentiating X^N Where N Is A Negative Integer - Video - Negative powers differentiation Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), you extend the power rule \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) to negative integer powers. The trick is to rewrite a reciprocal power such as \(\dfrac{1}{x^{k}}\) as \(x^{-k}\) first, then differentiate. This page shows the rewrite, the exponent arithmetic, and how to find the gradient of a reciprocal-power function.
A reciprocal power like \(\dfrac{1}{x^{k}}\) can be written with a negative index as \(x^{-k}\). Once it is in the form \(x^{n}\), the ordinary power rule applies exactly as it does for positive powers.
The power rule \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) holds for every non-zero integer \(n\), including negative ones. You multiply by the old index and then reduce the index by one; for a negative index, reducing by one makes it more negative (for example \(-3\) becomes \(-4\)).
Because differentiation is linear, a sum or difference of reciprocal-power terms is differentiated term by term. A single fraction over \(x^{k}\) is first split into separate power terms.
Rewrite a reciprocal power with a negative index:
Then apply the power rule (valid for every non-zero integer \(n\)):
How to differentiate a reciprocal power
- Rewrite: replace each \(\dfrac{1}{x^{k}}\) with \(x^{-k}\); split any single fraction over \(x^{k}\) into separate power terms first.
- Apply the power rule: multiply by the index and subtract one from it, writing the exponent arithmetic (\(n-1\)) on its own line.
- Tidy: if the question uses fraction form, rewrite the negative powers back as fractions \(x^{-m}=\dfrac{1}{x^{m}}\).
Rewrite with a negative index:
| \(y\) | \(=\) | \(\dfrac{1}{x^{3}}\) |
| \(=\) | \(x^{-3}\) |
Apply the power rule \(n\,x^{n-1}\) with \(n=-3\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(-3\,x^{-3-1}\) |
| \(=\) | \(-3\,x^{-4}\) |
Write the answer back in fraction form:
| \(\dfrac{dy}{dx}\) | \(=\) | \(-\dfrac{3}{x^{4}}\) |
\(\dfrac{dy}{dx}=-\dfrac{3}{x^{4}}\).
Rewrite each term with a negative index:
| \(f(x)\) | \(=\) | \(2x^{-2}-5x^{-1}\) |
Differentiate term by term:
| \(f'(x)\) | \(=\) | \(2(-2)x^{-2-1}-5(-1)x^{-1-1}\) |
| \(=\) | \(-4x^{-3}+5x^{-2}\) |
Back to fraction form:
| \(f'(x)\) | \(=\) | \(-\dfrac{4}{x^{3}}+\dfrac{5}{x^{2}}\) |
\(f'(x)=-\dfrac{4}{x^{3}}+\dfrac{5}{x^{2}}\).
Split the quotient over \(x^{3}\), then use negative indices:
| \(f(x)\) | \(=\) | \(\dfrac{x^{2}}{x^{3}}-\dfrac{4}{x^{3}}\) |
| \(=\) | \(x^{-1}-4x^{-3}\) |
Differentiate term by term:
| \(f'(x)\) | \(=\) | \(-1\,x^{-1-1}-4(-3)x^{-3-1}\) |
| \(=\) | \(-x^{-2}+12x^{-4}\) |
Back to fraction form:
| \(f'(x)\) | \(=\) | \(-\dfrac{1}{x^{2}}+\dfrac{12}{x^{4}}\) |
\(f'(x)=-\dfrac{1}{x^{2}}+\dfrac{12}{x^{4}}\).
Rewrite and differentiate:
| \(y\) | \(=\) | \(x^{-1}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(-1\,x^{-1-1}\) |
| \(=\) | \(-x^{-2}=-\dfrac{1}{x^{2}}\) |
Set the gradient equal to \(-\dfrac14\) and solve:
| \(-\dfrac{1}{x^{2}}\) | \(=\) | \(-\dfrac{1}{4}\) |
| \(x^{2}\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(\pm 2\) |
Find each \(y\)-coordinate from \(y=\dfrac1x\):
| \(y\) | \(=\) | \(\dfrac{1}{2}\ \text{ or }\ -\dfrac{1}{2}\) |
Gradient \(-\dfrac14\) at \(\left(2,\tfrac12\right)\) and \(\left(-2,-\tfrac12\right)\).
Common pitfalls
Frequently asked questions
How do you differentiate 1 over x cubed?
Rewrite \(\dfrac{1}{x^{3}}\) as \(x^{-3}\), then apply the power rule: \(-3x^{-4}=-\dfrac{3}{x^{4}}\).
Does the power rule work for negative powers?
Yes. \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) holds for every non-zero integer \(n\), including negative ones.
Why do you rewrite the fraction as a negative power?
The power rule only applies to something in the form \(x^{n}\). Writing \(\dfrac{1}{x^{k}}=x^{-k}\) puts it in that form so you can differentiate.
What does the index become after differentiating x to the minus three?
It becomes \(-4\), because you subtract one: \(-3-1=-4\). The derivative is \(-3x^{-4}\).
How do you differentiate a fraction like (x squared minus 4) over x cubed?
Split it first: \(\dfrac{x^{2}-4}{x^{3}}=x^{-1}-4x^{-3}\), then differentiate each term to get \(-x^{-2}+12x^{-4}\).