Differentiation finds dy/dx, the gradient function of a curve, which gives the gradient at any point. The power rule says that if y = xⁿ, then dy/dx = nxⁿ⁻¹: bring the power down to the front and reduce it by one. It works for any rational power, so roots and fractions must be rewritten as powers of x first.
Differentiation rules to know
- y = xⁿ gives dy/dx = nxⁿ⁻¹
- y = axⁿ gives dy/dx = anxⁿ⁻¹
- The derivative of a constant is 0
- Differentiate a sum or difference term by term
Useful rewrites
- √x = x^(1/2) and ∛x = x^(1/3)
- 1/xᵏ = x^(−k) and 1/√x = x^(−1/2)
What does differentiation tell you?
The derivative dy/dx is a formula for the gradient of a curve at any point. Substitute an x-value into it and you get the gradient of the curve, which is the gradient of the tangent, at that point. A positive value means the curve is rising there, a negative value means it is falling, and zero means it is momentarily flat.
The power rule
For any term xⁿ, bring the power down to multiply the front and then reduce the power by one. Constant multipliers stay where they are, and a sum is differentiated term by term. A constant on its own differentiates to 0, because a horizontal line has zero gradient: a “+5” simply disappears.
For example, 3x⁴ becomes 4 × 3x³ = 12x³, and −2x² becomes −4x.
Negative and fractional indices
The power rule works for any rational index, but only once each term is written as a power of x. Rewrite roots and reciprocals first: √x = x^(1/2), 1/x² = x^(−2), and 1/√x = x^(−1/2).
Take care reducing the index by one: −2 − 1 = −3, and ½ − 1 = −½. These subtractions are where most slips happen.
Rewrite products and quotients first
At this stage there is no product or quotient rule, so a product must be expanded and a quotient split into separate powers of x before you differentiate. For example, x²(2x + 5) becomes 2x³ + 5x², and (x² − 4)/x becomes x − 4x⁻¹.
You cannot differentiate a fraction by differentiating the top and the bottom separately.
The gradient at a point, and solving for a gradient
To find the gradient at a particular point, differentiate first, then substitute the x-value into dy/dx, not into y. The original equation gives the height of the curve; the derivative gives its gradient.
To find where the curve has a given gradient, set dy/dx equal to that value and solve for x. Setting dy/dx = 0 finds where the curve is flat, which is the first step towards stationary points.
Worked examples
Example 1. Find dy/dx for y = 3x⁴ − 2x² + 5.
- 3x⁴: bring the 4 down and reduce the power, giving 12x³.
- −2x²: 2 × (−2)x = −4x.
- +5: the derivative of a constant is 0.
Answer: dy/dx = 12x³ − 4x.
Marks: M1 for reducing a power by one, A1 for 12x³, A1 for −4x with the constant gone.
Example 2. Find dy/dx for y = √x + 1/x².
- Rewrite as powers: y = x^(1/2) + x^(−2).
- x^(1/2) differentiates to ½x^(−1/2).
- x^(−2) differentiates to −2x^(−3).
Answer: dy/dx = ½x^(−1/2) − 2x^(−3), which is 1/(2√x) − 2/x³.
Marks: M1 for rewriting, A1 for each term.
Example 3. Find dy/dx for y = x²(2x + 5).
- Expand first: y = 2x³ + 5x².
- Differentiate term by term.
Answer: dy/dx = 6x² + 10x.
Marks: M1 for expanding, A1 for the derivative.
Example 4. Find the gradient of y = x³ − 3x² + 2 at x = 1.
- Differentiate: dy/dx = 3x² − 6x.
- Substitute x = 1: 3(1) − 6(1) = −3.
Answer: The gradient is −3.
Marks: M1 for 3x² − 6x, A1 for −3.
Example 5. Find the x-values where y = x³ − 12x + 5 has gradient 0.
- dy/dx = 3x² − 12.
- Set 3x² − 12 = 0, so x² = 4.
- x = 2 or x = −2.
Answer: x = 2 or x = −2.
Marks: M1 for 3x² − 12 = 0, A1 for each value.
Common mistakes
- Substituting into y instead of dy/dx. The gradient comes from the derivative. Substituting into y gives the height of the curve.
- Not rewriting roots and fractions. Write √x as x^(1/2) and 1/x² as x^(−2) before using the power rule.
- Errors with negative indices. −2 − 1 = −3, not −1. Reduce the index carefully.
- Differentiating a fraction top and bottom. Split it into separate powers of x first.
- Keeping the constant. A constant term differentiates to 0.
How basic differentiation is examined
Differentiation starts in Pure 1 (WMA11) of Edexcel International A-Level, and in the first year of Edexcel A-Level (9MA0). Exam questions on this topic typically:
- ask you to write an expression as a sum of powers of x, then differentiate
- ask for the gradient of a curve at a given point
- ask where a curve has a particular gradient, often zero
- lead into tangents, normals and stationary points
The Pure 1 differentiation exam questions include all of these with mark schemes. For extra practice, try the Pure 1 differentiation worksheet or the A-Level differentiation revision engine.
To see why the power rule works, look at differentiation from first principles. Next comes tangents and normals. If indices need a refresh, revisit indices and surds.
Frequently asked questions
What is the power rule in differentiation?
If y = xⁿ, then dy/dx = nxⁿ⁻¹. Multiply by the power, then reduce the power by one.
How do you differentiate √x?
Write it as x^(1/2), then use the power rule: the derivative is ½x^(−1/2), or 1/(2√x).
What does dy/dx mean?
It is the rate of change of y with respect to x: the gradient of the curve y = f(x) at any point.