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Differentiating powers and roots

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The power rule

The rule that underlies everything:

$\frac{d}{dx}\left(x^n\right) = nx^{n-1}$

In words: multiply by the power, then reduce the power by one.
It holds for every real $n$ — positive, negative, whole or fractional.
With a coefficient in front, the coefficient just comes along:

$\frac{d}{dx}\left(ax^n\right) = anx^{n-1}$

The two special cases

A constant differentiates to zero. $\dfrac{d}{dx}(7) = 0$ — a horizontal line has gradient 0.
$x$ differentiates to 1. Writing $x = x^1$ , the rule gives $1 \times x^0 = 1$ .

Sums and differences

Differentiate term by term. The derivative of a sum is the sum of the derivatives.

$\frac{d}{dx}\left(3x^4 - 5x^2 + 7x - 9\right) = 12x^3 - 10x + 7$

Negative powers

A term in a denominator becomes a negative power, and then the ordinary rule applies.
Rewrite first, differentiate second.

$\frac{1}{x^3} = x^{-3} \quad \Rightarrow \quad \frac{d}{dx}\left(x^{-3}\right) = -3x^{-4} = -\frac{3}{x^4}$

Watch the arithmetic on the exponent. $-3 - 1 = -4$ , and the power decreases, so it becomes more negative.

Fractional powers — roots

Roots become fractional powers:
- $\sqrt{x} = x^{1/2}$
- $\sqrt[3]{x} = x^{1/3}$
- $\dfrac{1}{\sqrt{x}} = x^{-1/2}$
The 2025 assessor report lists "successfully wrote surd expressions into fractional index form, before differentiating" as an Achieved behaviour, and its absence — "did not recognise that $\sqrt{x}$ is $x^{1/2}$ " — among the Not Achieved. Converting is not optional.
Worked through:

$\frac{d}{dx}\left(\sqrt{x}\right) = \frac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

$\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) = \frac{d}{dx}\left(x^{-1/2}\right) = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x\sqrt{x}}$

Note the subtraction: $\frac{1}{2} - 1 = -\frac{1}{2}$ , and $-\frac{1}{2} - 1 = -\frac{3}{2}$ .

Preparing an expression before differentiating

The power rule works on terms, not on products or quotients. If an expression is written as one, rewrite it as a sum of terms first where you can.
Expand a bracket:

$y = x^2(x + 3) = x^3 + 3x^2 \quad \Rightarrow \quad \frac{dy}{dx} = 3x^2 + 6x$

Split a fraction with a single term underneath:

$y = \frac{x^3 + 4x}{x^2} = x + 4x^{-1} \quad \Rightarrow \quad \frac{dy}{dx} = 1 - 4x^{-2}$

Note that you cannot split a fraction across the denominator. $\dfrac{1}{x + 2}$ is not $\dfrac{1}{x} + \dfrac{1}{2}$ ; that one needs the chain or quotient rule.

Notation

These all mean the same thing:
- $\dfrac{dy}{dx}$ — used when the function is written as $y = \ldots$
- $f'(x)$ — used when the function is written as $f(x) = \ldots$
- $\dfrac{d}{dx}\left[\;\right]$ — an instruction to differentiate what follows
The second derivative is $\dfrac{d^2y}{dx^2}$ or $f''(x)$ , found by differentiating again.

Higher derivatives

Differentiate repeatedly, one step at a time:

$y = x^5 \;\to\; \frac{dy}{dx} = 5x^4 \;\to\; \frac{d^2y}{dx^2} = 20x^3 \;\to\; \frac{d^3y}{dx^3} = 60x^2$

Tips and Tricks

Convert every root and every reciprocal to an index before you differentiate. Almost every "could not differentiate" mark in the assessor reports traces back to a surd or a fraction that was never converted. Write $\sqrt[3]{x^2}$ as $x^{2/3}$ on its own line, then differentiate.

The exponent decreases by 1. For negative powers that makes it more negative: $x^{-2}$ gives $-2x^{-3}$ , not $-2x^{-1}$ . This is the single most common index slip in the paper.

Expand brackets and split single-denominator fractions first. It converts a product or quotient problem into a power-rule problem and avoids the harder rules altogether — fewer steps, fewer errors.

$-x^2$ means $-(x^2)$ , not $(-x)^2$ . The report lists "misinterpreted expressions such as $-x^2$ " among Not Achieved behaviours at Level 2, and the same slip persists at Level 3.

If the question says "you do not need to simplify", hand in $\frac{1}{2}x^{-1/2}$ and stop. The report is emphatic that candidates "continue to simplify their derivatives and make careless mistakes when doing so, resulting in them losing their grade in the question".

Worked ExamplePreparing before differentiating

Find $\dfrac{dy}{dx}$ for $y = \dfrac{2x^3 - \sqrt{x}}{x^2}$ .

Step 1 — Recognise that the power rule does not apply directly

This is a quotient, so the power rule cannot be used on it as written. But the denominator is a single term, so the fraction can be split — which is far easier than the quotient rule.

Step 2 — Convert the surd to an index

Before splitting, rewrite $\sqrt{x}$ :

$y = \frac{2x^3 - x^{1/2}}{x^2}$

Step 3 — Split the fraction

Divide each term in the numerator by $x^2$ :

$y = \frac{2x^3}{x^2} - \frac{x^{1/2}}{x^2}$

Step 4 — Simplify each term using index laws

Dividing subtracts the exponents:

$\frac{2x^3}{x^2} = 2x^{3-2} = 2x$

$\frac{x^{1/2}}{x^2} = x^{1/2 - 2} = x^{-3/2}$

So:

$y = 2x - x^{-3/2}$

Step 5 — Differentiate term by term

First term: $\dfrac{d}{dx}(2x) = 2$ .

Second term: multiply by the power and reduce it by 1. The power is $-\frac{3}{2}$ , so the new power is $-\frac{3}{2} - 1 = -\frac{5}{2}$ :

$\frac{d}{dx}\left(-x^{-3/2}\right) = -\left(-\frac{3}{2}\right)x^{-5/2} = \frac{3}{2}x^{-5/2}$

Step 6 — Combine

$\frac{dy}{dx} = 2 + \frac{3}{2}x^{-5/2}$

Step 7 — Present in surd form if required

$\frac{dy}{dx} = 2 + \frac{3}{2x^2\sqrt{x}}$

Check the sign. The original second term was $-x^{-3/2}$ , and differentiating a negative power gives a negative coefficient; two negatives make the $+\frac{3}{2}$ ✓