How to Differentiate: The Power Rule (IGCSE Maths)
Learn how to differentiate functions using the power rule — one of the most important skills in IGCSE Maths. This page covers the gradient of a curve, the gradient function dy/dx, differentiating single terms and sums, and how to handle negative and fractional powers. Clear worked examples and a randomised practice room with auto-marked questions help you build speed and confidence.
Make sure you are comfortable with Gradient of a Straight Line, Plotting Straight-Line Graphs, and Index Notation before starting.
The Gradient of a Curve
For a straight line, the gradient is constant: the same at every point. A curve is different. Its gradient changes at every point, so you need a method that gives the exact gradient at any chosen point. That method is called differentiation.
Straight Line vs Curve
Straight line, same gradient everywhere
Curve: gradient changes at every point
The gradient of a curve at any point is the gradient of the tangent at that point. Differentiation gives you the exact gradient without drawing. The result is called the gradient function, written \(\dfrac{dy}{dx}\).
The gradient of a curve changes at every point. You need calculus to find it exactly.
Differentiating a function gives the gradient function \(\dfrac{dy}{dx}\). Substitute any \(x\) to find the gradient there.
Positive \(\dfrac{dy}{dx}\): curve rising. Negative: curve falling. Zero: maximum or minimum point.
The gradient at a point is the slope of the tangent line there. Differentiation finds this exactly.
The Power Rule
The rule for differentiating any power of \(x\) is straightforward: multiply by the power, then reduce the power by 1.
If \(y = kx^n\), then:
\[\frac{dy}{dx} = nkx^{n-1}\]Multiply by the power, then reduce the power by 1.
If \(y = kx\), then \(\dfrac{dy}{dx} = k\).
If \(y = k\) (a constant), then \(\dfrac{dy}{dx} = 0\).
A constant has no slope, so its derivative is always zero.
Differentiate each term separately:
\[y = 3x^3 + 2x^2 - 5\] \[\frac{dy}{dx} = 9x^2 + 4x\]The constant \(-5\) disappears.
| Function \(y\) | Gradient function \(\dfrac{dy}{dx}\) | Gradient at \(x = 2\) |
|---|---|---|
| \(x^2\) | \(2x\) | \(4\) |
| \(x^3\) | \(3x^2\) | \(12\) |
| \(4x^3\) | \(12x^2\) | \(48\) |
| \(x^{-1} = \dfrac{1}{x}\) | \(-x^{-2} = -\dfrac{1}{x^2}\) | \(-\dfrac{1}{4}\) |
| \(x^{1/2} = \sqrt{x}\) | \(\dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}\) | \(\dfrac{1}{2\sqrt{2}}\) |
| \(7\) (constant) | \(0\) | \(0\) |
Before you differentiate: rewrite first
Rewrite using index notation, then apply the power rule normally.
Rewrite using index notation, then apply the power rule normally.
You cannot differentiate a product directly. Always expand and simplify first.
💡 Example 1: Basic power
Differentiate \(y = 5x^3\).
\[\frac{dy}{dx} = 3 \times 5x^{3-1} = 15x^2\]💡 Example 2: Negative power
Differentiate \(y = \dfrac{4}{x^2} = 4x^{-2}\).
\[\frac{dy}{dx} = -2 \times 4x^{-3} = -\frac{8}{x^3}\]💡 Example 3: Sum of terms
Differentiate \(y = 3x^4 - 6x^2 + 5\).
\[\frac{dy}{dx} = 12x^3 - 12x\]The constant \(5\) disappears; its gradient is zero.
💡 Example 4: Expand brackets first
Differentiate \(y = (2x+1)(x-3)\).
Expand: \(y = 2x^2 - 5x - 3\)
\[\frac{dy}{dx} = 4x - 5\]🔑 Key Points
- Always rewrite in index form before differentiating (roots, fractions).
- Expand brackets first; you cannot differentiate a product directly.
- Constants differentiate to zero.
- Differentiate each term in a sum separately.
⚠️ Common Mistakes
- Forgetting to rewrite \(\dfrac{1}{x}\) as \(x^{-1}\) before applying the rule.
- Trying to differentiate brackets without expanding first.
- Losing the constant: \(\dfrac{d}{dx}(5) = 0\), not \(5\).
- Getting the sign wrong with negative powers: \(-1 \times -2 = +2\).