Maths › Differentiation › The derivative from first principles
The derivative from first principles
A curve has no single gradient, but each point on it does, and catching that number takes a genuinely new idea: measure the slope of a chord, then let the chord shrink. What survives the shrinking is the derivative, and everything else in calculus is built on top of this one move.
Builds on Straight lines.
IN THIS TOPIC
- Explain the derivative as the limit of chord gradients, and use f'(x) and dy/dx notation.
- Differentiate small powers of x from first principles.
- Sketch a gradient function from the graph of a curve.
- Read the second derivative as the rate of change of the gradient.
COMMON MISCONCEPTION
The gradient of a curve at a point is the gradient of the line to a nearby point.
The shrinking chord
A straight line has one gradient. A curve does not. So start with something you can actually measure, the gradient of a chord joining (x, f(x)) to a neighbouring point (x + h, f(x + h)), which is rise over run as ever. The chord's gradient is not the answer. It is an approximation whose error shrinks as h shrinks, and the gradient of the curve is the number those approximations close in on.
That limit is the derivative, written f'(x) or dy/dx, and it gives the gradient of the tangent at the point. It is printed in the formulae booklet, under Differentiation in the A level Mathematics section, so you can copy the statement straight off the page and spend your time on the algebra instead. Computing it straight from the limit is called differentiating from first principles. Some papers use δx where this page uses h. Same quantity, same proof, so do not let the swap unsettle you.
WORKED EXAMPLE
x² from first principles
Prove from first principles that the derivative of f(x) = x2 is 2x.
Form the chord gradient: [(x + h)2 − x2]/h = (2xh + h2)/h.
For h ≠ 0, divide through, and the chord gradient is 2x + h.
As h → 0 the surviving part is 2x, so f'(x) = 2x. ∎
Cancel the h before you take the limit. Substituting h = 0 into the original fraction gives 0/0, and dodging that is the whole reason the algebra is there.
GUIDED PRACTICE
A cubic from first principles
Differentiate f(x) = x3 from first principles, before opening the working.
Show the working
Expand: (x + h)3 = x3 + 3x2h + 3xh2 + h3.
The chord gradient is (3x2h + 3xh2 + h3)/h = 3x2 + 3xh + h2.
As h → 0, both terms carrying h vanish, leaving f'(x) = 3x2. ∎
Every surviving term of the expansion had exactly one h to cancel. The terms with more of them die in the limit.
The gradient is a function
Read the slope at every point of a curve and a whole new function drops out, the gradient function. The two graphs then hold a conversation. Where the curve climbs, the gradient function sits above the axis; where the curve is momentarily flat, the gradient function crosses zero.
Sketching y = f'(x) from y = f(x) is a standard four-mark question and the order of work decides whether you get them. Mark the zeros first, directly under the flat points. Then fix the sign of the gradient function on each stretch between them, and only then worry about how steep the thing looks.
INDEPENDENT PRACTICE
First principles with two terms
Differentiate f(x) = x2 − 4x from first principles.
Show the working
The chord gradient is [(x + h)2 − 4(x + h) − x2 + 4x]/h.
The numerator expands to 2xh + h2 − 4h, so the chord gradient is 2x + h − 4.
As h → 0, f'(x) = 2x − 4. ∎
The −4x term contributed −4, its own gradient, and the limit dealt with each term on its own. Differentiation will always respect sums this way.
Differentiating twice
Differentiate the gradient function itself and you get the second derivative, written f''(x) or d2y/dx2. It measures how fast the gradient is changing. On a stretch where the curve bends upwards the gradient is increasing, so f'' is positive there, and the next lesson turns that single observation into the standard test for hilltops and valleys. Everything downstream, tangents, turning points, optimisation, rests on the limit at the top of this page.
ASSESSMENT FOCUS
- First-principles proofs are marked line by line. Chord gradient stated, expansion shown, h cancelled, limit taken. Skip the cancellation and the middle marks go with it.
- Write the limit statement as well as the algebra. “As h → 0, the gradient → 2x” is the concluding mark.
- f'(x), dy/dx and “the gradient function” all name the same object, and a question stem may use any of the three.
- The chord gradient before the limit, 2x + h, is exact for that chord. Only the limit turns it into a tangent gradient.
CHECK YOURSELF
From first principles, find the gradient of y = 3x2 at the point (2, 12).
Show a hint
Chord from x = 2, expand, cancel h, then let h → 0.
Show the answer
The chord gradient is [3(2 + h)2 − 12]/h = (12h + 3h2)/h = 12 + 3h.
As h → 0 the gradient approaches 12.
The general derivative 6x, evaluated at x = 2, agrees. Having two routes to the same number is how you check first-principles work.
The derivative is the limit of chord gradients as the chord shrinks to a point.
Cancel h first, then let it vanish; what survives is the gradient function.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the derivative from first principles questions page.
CHECK YOUR PROGRESS
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- Explain the derivative as the limit of chord gradients, and use f'(x) and dy/dx notation.
- Differentiate small powers of x from first principles.
- Sketch a gradient function from the graph of a curve.
- Read the second derivative as the rate of change of the gradient.
Open the full revision checklist to see every objective in the course in one place.