Differentiating Rational Powers
Learn to differentiate rational powers for Queensland Year 11 Mathematical Methods (QCAA). The power rule also works for fractional indices, so surds and roots can be differentiated once they are rewritten in index form.
You will learn to rewrite square roots and cube roots as fractional powers, apply the power rule to the fractional index, tidy the answer back into surd form, and find exact gradients on curves involving roots.
Every question with a fully worked solution.
- Differentiating Rational Powers - Video - Fractional powers differentiation | Derivative rules Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), the power rule \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) also works for rational (fractional) powers. The key first step is to rewrite surds and roots as fractional indices — \(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{\sqrt{x}}=x^{-1/2}\), \(\sqrt[3]{x^{2}}=x^{2/3}\) — then apply the power rule. This page shows the rewrite, the index arithmetic, and finding exact gradients.
A rational power is a power with a fraction as its index, such as \(x^{1/2}\) or \(x^{2/3}\). Every surd or root can be written this way: \(\sqrt{x}=x^{1/2}\), \(\sqrt[3]{x}=x^{1/3}\), and \(\sqrt[n]{x^{m}}=x^{m/n}\).
Once a root is written as a fractional index, the ordinary power rule \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) applies. You multiply by the fractional index and subtract one from it — for instance \(\tfrac12-1=-\tfrac12\).
Sums of rational and integer powers are differentiated term by term, and answers are usually written back in surd form, often after rationalising a denominator.
Rewrite roots as fractional indices:
Then apply the power rule (valid for rational \(n\)):
How to differentiate a rational power
- Rewrite: convert every surd or root to a fractional index (\(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{\sqrt{x}}=x^{-1/2}\)).
- Apply the power rule: multiply by the fractional index and subtract one from it, showing the arithmetic \(\tfrac{m}{n}-1\) on its own line.
- Return to surd form: rewrite the negative or fractional power as a surd, and rationalise the denominator if needed.
Rewrite the root as a fractional index:
| \(y\) | \(=\) | \(\sqrt{x}\) |
| \(=\) | \(x^{1/2}\) |
Apply the power rule \(n\,x^{n-1}\) with \(n=\tfrac12\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(\tfrac12\,x^{\tfrac12-1}\) |
| \(=\) | \(\tfrac12\,x^{-1/2}\) |
Return to surd form:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{1}{2\sqrt{x}}\) |
\(\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\).
Rewrite with a negative fractional index:
| \(f(x)\) | \(=\) | \(\dfrac{1}{\sqrt{x}}\) |
| \(=\) | \(x^{-1/2}\) |
Apply the power rule with \(n=-\tfrac12\):
| \(f'(x)\) | \(=\) | \(-\tfrac12\,x^{-\tfrac12-1}\) |
| \(=\) | \(-\tfrac12\,x^{-3/2}\) |
Return to surd form:
| \(f'(x)\) | \(=\) | \(-\dfrac{1}{2\,x^{3/2}}\) |
| \(=\) | \(-\dfrac{1}{2x\sqrt{x}}\) |
\(f'(x)=-\dfrac{1}{2x\sqrt{x}}\).
Rewrite the root as a fractional index:
| \(f(x)\) | \(=\) | \(x^{2}+4x^{1/2}\) |
Differentiate term by term:
| \(f'(x)\) | \(=\) | \(2x+4\cdot\tfrac12\,x^{\tfrac12-1}\) |
| \(=\) | \(2x+2x^{-1/2}\) |
Return to surd form:
| \(f'(x)\) | \(=\) | \(2x+\dfrac{2}{\sqrt{x}}\) |
\(f'(x)=2x+\dfrac{2}{\sqrt{x}}\).
Rewrite and differentiate:
| \(y\) | \(=\) | \(x^{1/2}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt{x}}\) |
Substitute \(x=5\):
| \(\left.\dfrac{dy}{dx}\right|_{x=5}\) | \(=\) | \(\dfrac{1}{2\sqrt{5}}\) |
Rationalise the denominator:
| \(=\) | \(\dfrac{1}{2\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\) | |
| \(=\) | \(\dfrac{\sqrt{5}}{10}\) |
Exact gradient at \(x=5\) is \(\dfrac{\sqrt{5}}{10}\).
Common pitfalls
Frequently asked questions
How do you differentiate the square root of x?
Write \(\sqrt{x}=x^{1/2}\), then apply the power rule: \(\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt{x}}\).
Does the power rule work for fractional powers?
Yes. \(\dfrac{d}{dx}\!\left(x^{n}\right)=n\,x^{n-1}\) holds for rational \(n\), so \(x^{2/3}\) differentiates to \(\tfrac23 x^{-1/3}\).
How do you differentiate 1 over the square root of x?
Write it as \(x^{-1/2}\), then differentiate: \(-\tfrac12 x^{-3/2}=-\dfrac{1}{2x\sqrt{x}}\).
How do you differentiate a cube root?
Rewrite it as a fractional index: \(\sqrt[3]{x^{2}}=x^{2/3}\), then apply the power rule to get \(\tfrac23 x^{-1/3}\).
Why is my rational-power answer left as a surd?
Because exact form is expected. Rewrite negative fractional powers back as surds, for example \(x^{-1/2}=\dfrac{1}{\sqrt{x}}\), and rationalise if there is a surd in the denominator.