Differentiation from first principles finds the derivative as a limit. The gradient of the chord from x to x + h is (f(x + h) − f(x))/h, and as h → 0 the chord becomes the tangent, so f′(x) = lim (f(x + h) − f(x))/h as h → 0. Expand, simplify until the h on the bottom cancels, then let h tend to zero.
The first principles definition
- Gradient of a chord: (f(x + h) − f(x)) / h
- Derivative: f′(x) = lim as h → 0 of (f(x + h) − f(x)) / h
- (x + h)² = x² + 2xh + h²
- (x + h)³ = x³ + 3x²h + 3xh² + h³
Simplify until the h on the denominator cancels, then let h → 0. Putting h = 0 at the start gives 0/0.
What does first principles mean?
The power rule is fast, but first principles shows where it comes from. Take a point on the curve at x and a second point a small step h further along, at x + h. The chord joining them has gradient (f(x + h) − f(x))/h: the change in y over the change in x.
As h gets smaller, the chord lies closer and closer to the curve, and its gradient approaches the gradient of the tangent. The derivative is defined as that limit.
Why can't you just put h = 0?
Substituting h = 0 straight away gives 0/0, which is undefined. The method is to simplify the chord gradient first, so that the h in the denominator cancels with a factor of h in the numerator. Only then do you let h tend to zero: any term that still contains h disappears.
First principles for x² and quadratics
For f(x) = x², f(x + h) = x² + 2xh + h². Subtracting f(x) leaves 2xh + h². Dividing by h gives 2x + h, and as h → 0 this tends to 2x. The same routine works for any quadratic ax² + bx + c and gives 2ax + b.
Every x² term on the top cancels, and one factor of h cancels with the bottom. If the h will not cancel, recheck the expansion.
First principles for x³, and the power rule
Expanding (x + h)³ = x³ + 3x²h + 3xh² + h³, the x³ cancels, one h divides out to leave 3x² + 3xh + h², and the limit leaves 3x². So x² differentiates to 2x and x³ to 3x², matching the power rule xⁿ → nxⁿ⁻¹. First principles is exactly what proves it.
First principles for 1/x
For f(x) = 1/x, combine the two fractions over a common denominator before dividing by h: 1/(x + h) − 1/x = −h / (x(x + h)). Dividing by h gives −1/(x(x + h)), which tends to −1/x² as h → 0. That agrees with the power rule on x⁻¹.
Worked examples
Example 1. Find the gradient of the chord on y = x² joining x = 2 and x = 2.1.
- Heights: f(2.1) = 4.41 and f(2) = 4.
- Gradient = (4.41 − 4)/(2.1 − 2) = 0.41/0.1 = 4.1.
Answer: 4.1, close to the true gradient of 4 at x = 2.
Marks: M1 for the chord gradient, A1 for 4.1.
Example 2. Use first principles to find f′(x) for f(x) = 2x² − 3x + 1.
- f(x + h) = 2(x + h)² − 3(x + h) + 1 = 2x² + 4xh + 2h² − 3x − 3h + 1.
- Subtract f(x): the top becomes 4xh + 2h² − 3h.
- Divide by h: 4x + 2h − 3.
- Let h → 0.
Answer: f′(x) = 4x − 3.
Marks: M1 for expanding 2(x + h)², M1 for cancelling h, A1 for 4x − 3.
Example 3. Use first principles to differentiate f(x) = x³ + x.
- f(x + h) = x³ + 3x²h + 3xh² + h³ + x + h.
- Subtract f(x): 3x²h + 3xh² + h³ + h.
- Divide by h: 3x² + 3xh + h² + 1.
- Let h → 0.
Answer: f′(x) = 3x² + 1.
Marks: M1 for expanding (x + h)³, M1 for cancelling h, A1 for 3x² + 1.
Example 4. Use first principles to find f′(x) for f(x) = x², then the gradient at x = 3.
- ((x + h)² − x²)/h = (2xh + h²)/h = 2x + h, which tends to 2x.
- So f′(x) = 2x, and at x = 3 the gradient is 6.
Answer: The gradient at x = 3 is 6.
Marks: M1 for f′(x) = 2x, A1 for 6.
Common mistakes
- Putting h = 0 too early. Simplify until the h on the denominator cancels first.
- Expanding (x + h)² as x² + h². (x + h)² = x² + 2xh + h². The middle term is the one that survives.
- Not writing the limit. Show “as h → 0” clearly; the final mark depends on it.
- Forgetting to subtract f(x). The numerator is f(x + h) − f(x), not just f(x + h).
- Dividing by h before combining fractions. For 1/x, combine 1/(x + h) − 1/x into a single fraction first.
How first principles is examined
First principles is part of Pure 1 (WMA11) in Edexcel International A-Level and of first-year Pure in Edexcel A-Level (9MA0). It is a proof topic, so method marks reward clear working. You may be asked to:
- expand f(x + h) for a given polynomial
- show that the chord gradient simplifies to a stated expression
- use the limit as h → 0 to prove a derivative
- then use that derivative to find a gradient at a point
You can find full first principles questions with mark schemes among the Pure 1 differentiation exam questions, and more in the differentiation worksheet. The differentiation revision engine is useful once you move on to the power rule.
Once the idea is clear, the basic differentiation course uses the power rule that first principles proves.
Frequently asked questions
What is the first principles formula?
f′(x) = lim as h → 0 of (f(x + h) − f(x))/h. It is the gradient of a chord as the chord shrinks to a single point.
Do I need first principles in the exam?
Yes. You can be asked to prove the derivative of a polynomial from first principles, usually a quadratic or cubic, and the method is marked step by step.
Why does first principles give the gradient of the tangent?
As h → 0, the second point slides along the curve towards the first, and the chord becomes the tangent at that point.