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Additional Mathematics · Lesson

Differentiate fractional and negative powers

The rule for x to the power n is easy, until the question hands you a root or a fraction.

On this page
  1. How do you turn roots and fractions into powers?
  2. Worked example
  3. The mistake to watch for
  4. Check yourself
  5. Where this leads next

To differentiate any term axn, multiply by the power and then reduce the power by one: d/dx of axn = anxn−1. This holds when n is negative or a fraction as well. The first job is to rewrite roots and fractions as powers of x.

This is the base skill of differentiation techniques. Every later rule finishes by applying it.

How do you turn roots and fractions into powers?

Use the index laws first, then differentiate. The common conversions are:

Written asPower form
√xx1/2
1/xx−1
1/x²x−2
1/√xx−1/2
∛xx1/3
(x² + 3)/xx + 3x−1

A fraction with a single term underneath can always be split. Divide each term on top by the bottom, then simplify each one using the index laws.

Worked example

Differentiate y = 2x³ − 6√x + 4/x², and find the gradient at x = 4.

Step 1, rewrite: y = 2x³ − 6x1/2 + 4x−2.

Step 2, differentiate term by term:

  • 2x³ gives 6x².
  • −6x1/2 gives −6 × ½ × x−1/2 = −3x−1/2.
  • 4x−2 gives 4 × (−2) × x−3 = −8x−3.

So dy/dx = 6x² − 3x−1/2 − 8x−3 = 6x² − 3/√x − 8/x³.

Step 3, substitute x = 4: 6 × 16 = 96, then 3/√4 = 3/2 = 1.5, then 8/4³ = 8/64 = 0.125.

Gradient = 96 − 1.5 − 0.125 = 94.375.

The mistake to watch for

A common slip is to reduce the power from the wrong starting value, especially when it is already negative.

Mistaken working: y = 4/x² → dy/dx = 4 × (−2) × x−1 = −8/x

The student treated −2 − 1 as −1.

The power starts at −2, and reducing it by one gives −3, not −1. So the correct derivative is −8x−3 = −8/x³. Writing “new power = old power − 1” on your line, with the arithmetic shown, prevents this.

The same care applies to fractions: x1/2 becomes x−1/2, because ½ − 1 = −½.

Check yourself

1. Differentiate y = x3/2.

Show answer

Bring down 3/2 and reduce the power: 3/2 − 1 = 1/2.

dy/dx = (3/2)x1/2

2. Differentiate y = 5/√x and find the gradient at x = 4.

Show answer

Rewrite: y = 5x−1/2. Then dy/dx = 5 × (−½) × x−3/2 = −(5/2)x−3/2.

At x = 4, x3/2 = 8, so the gradient is −5/(2 × 8) = −5/16.

3. Differentiate y = (x² + 3)/x and find the gradient at x = 3.

Show answer

Split the fraction: y = x + 3x−1. Then dy/dx = 1 − 3x−2 = 1 − 3/x².

At x = 3: 1 − 3/9 = 1 − 1/3 = 2/3.

Where this leads next

With powers secure, move on to differentiating a product, where each factor is differentiated with this same rule. The calculus shape and rate explorer shows how the gradient formula relates to the curve, and the non-calculator working trainer builds the fluency for fractions and roots.

Students who know the rule but drop signs or indices under time pressure are the ones our teachers spend the most time with in online one-to-one Additional Mathematics tuition.

Questions people ask

Do I differentiate 1/x² using the quotient rule?

No. Rewrite it as x^(−2) and use the power rule: bring the −2 down and reduce the power by one, giving −2x^(−3), which is −2/x³. The quotient rule also works, but it is slower and easier to get wrong when the numerator is just a number.

What is the derivative of a constant?

It is zero, because a constant does not change as x changes. A graph of y = 7 is a horizontal line with gradient 0. So in a polynomial, a term with no x simply disappears when you differentiate.

Should I leave the answer with negative powers?

Follow the question. If it asks for the gradient at a value, substitute straight into the power form or the fraction form. If it asks you to simplify, give the answer with positive powers and roots, such as 3/√x rather than 3x^(−1/2).

Updated:

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