Calculus

Derivatives from First Principles: The Limit Definition

Understand the limit definition of the derivative and how instantaneous rate of change emerges from average rates. Master the first principles method with worked examples.

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Vectora Team
STEM Education
12 min read
2026-04-10

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What Is a Derivative?

The derivative of a function f(x)f(x) at a point x=ax = a measures the instantaneous rate of change of ff at that point. Geometrically, it gives the slope of the tangent line to the curve at (a,f(a))(a, f(a)).

The core idea: start with the average rate of change over an interval, then shrink the interval to zero.

Learning Goals: By the end of this guide, you should be able to:

  1. Explain the concept of a limit and how it relates to derivatives.
  2. Differentiate simple functions from first principles.
  3. Interpret the derivative as a rate of change and as a gradient.
  4. Recognise when a function is not differentiable at a point.

The First Principles Definition

The derivative of f(x)f(x) is defined as:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

The expression f(x+h)f(x)h\frac{f(x+h) - f(x)}{h} is the difference quotient — the slope of the secant line through (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)).

As h0h \to 0, the secant line pivots toward the tangent line, and the difference quotient approaches the derivative.


Worked Examples

Example 1: f(x)=x2f(x) = x^2

f(x)=limh0(x+h)2x2h=limh0x2+2xh+h2x2h=limh02xh+h2hf'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} =limh0(2x+h)=2x= \lim_{h \to 0} (2x + h) = 2x

So the gradient of y=x2y = x^2 at any point xx is 2x2x. At x=3x = 3, the tangent has slope 66.

Example 2: f(x)=1xf(x) = \frac{1}{x}

f(x)=limh01x+h1xh=limh0x(x+h)hx(x+h)=limh01x(x+h)=1x2f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{x - (x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2}

Example 3: f(x)=xf(x) = \sqrt{x}

f(x)=limh0x+hxhf'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}

Multiply by the conjugate x+h+xx+h+x\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}:

=limh0(x+h)xh(x+h+x)=limh01x+h+x=12x= \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}

From First Principles to Rules

The first principles method confirms the standard differentiation rules:

FunctionDerivativeObtained via first principles
xnx^nnxn1nx^{n-1}Binomial expansion
sinx\sin xcosx\cos xUsing limh0sinhh=1\lim_{h\to 0}\frac{\sin h}{h} = 1
exe^xexe^xUsing the definition of ee

Once proven, you can use these rules directly without re-deriving from limits each time.


When Does the Derivative Not Exist?

A function is not differentiable at a point where:

  1. There is a sharp corner — the left and right limits of the gradient disagree (e.g., x|x| at x=0x = 0).
  2. There is a vertical tangent — the gradient approaches infinity (e.g., x3\sqrt[3]{x} at x=0x = 0).
  3. There is a discontinuity — the function is not continuous at that point.

Common Mistakes

  1. Forgetting to take the limit — Computing f(x+h)f(x)h\frac{f(x+h) - f(x)}{h} and stopping there gives the difference quotient, not the derivative.
  2. Algebraic errors in expansion — Carefully expand (x+h)2(x + h)^2, (x+h)3(x + h)^3, etc. Missing a term is the most common source of error.
  3. Dividing by zero — You must algebraically cancel the hh in the denominator before substituting h=0h = 0.

Exam Tips (A-Level / AP / IB)

  • First principles questions almost always involve x2x^2, x3x^3, 1x\frac{1}{x} or x\sqrt{x}. Master these four.
  • Show every algebraic step — marks are awarded for the process, not just the final answer.
  • If the question says "from first principles" or "using the definition", you must use the limit formula. Shortcut rules earn zero marks.

Frequently Asked Questions

Why can't I just use the power rule directly?

You can — once it's been proven! But understanding the limit definition is essential because it's the foundation of all differentiation rules. Exam boards specifically test this understanding.

What is the relationship between continuity and differentiability?

Differentiability implies continuity: if f(a)f'(a) exists, then ff must be continuous at aa. But continuity does NOT imply differentiability — x|x| is continuous everywhere but not differentiable at x=0x = 0.


References & Further Reading

This article was created by the Vectora Editorial Team and is reviewed for alignment with AP, IB, and A-Level curricula. Content is based on standard academic sources in chemistry, physics, biology, and mathematics.

Published: 2026-04-10

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