Differentiation: Definition and Basic Rules

What you'll learn

This revision guide covers the foundational principles of differentiation tested in AP Calculus AB. You will master the limit definition of the derivative, understand what derivatives represent both geometrically and analytically, and apply the core differentiation rules with confidence. These techniques form the basis for all further calculus work you will encounter.

Key terms and definitions

Derivative — The instantaneous rate of change of a function at a point; the limit of the average rate of change as the interval approaches zero.

Differentiable — A function is differentiable at a point if its derivative exists at that point; the function must be continuous and have no sharp corners or vertical tangents there.

Power Rule — The differentiation rule stating that if f(x) = x^n, then f'(x) = nx^(n-1) for any real number n.

Chain Rule — The rule for differentiating composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) · g'(x).

Product Rule — The rule for differentiating the product of two functions: if y = uv, then dy/dx = u(dv/dx) + v(du/dx).

Quotient Rule — The rule for differentiating the quotient of two functions: if y = u/v, then dy/dx = [v(du/dx) - u(dv/dx)]/v².

Tangent line — A straight line that touches a curve at a single point and has the same gradient as the curve at that point; its equation is y - y₁ = f'(x₁)(x - x₁).

Normal line — A line perpendicular to the tangent line at a point on a curve; its gradient is the negative reciprocal of the derivative at that point.

Core concepts

The derivative from first principles

The derivative of a function f(x) at a point x is defined using the limit:

f'(x) = lim[h→0] [f(x + h) - f(x)]/h

This is also called the limit definition or derivative from first principles.

Alternative notations for the derivative include:

• f'(x) — Lagrange notation
• dy/dx — Leibniz notation
• df/dx — Leibniz alternative
• Df(x) — Operator notation

Geometric interpretation: The derivative represents the gradient (slope) of the tangent line to the curve y = f(x) at the point (x, f(x)).

Physical interpretation: The derivative represents the instantaneous rate of change of the dependent variable with respect to the independent variable. For position functions s(t), the derivative s'(t) gives velocity.

To differentiate from first principles:

1. Write f(x + h)
2. Substitute into [f(x + h) - f(x)]/h
3. Simplify the numerator algebraically
4. Cancel the factor of h
5. Evaluate the limit as h → 0

The power rule and basic derivatives

The power rule is the most fundamental differentiation rule:

If f(x) = x^n, then f'(x) = nx^(n-1)

This applies for all real values of n, including:

• Positive integers: x⁵ → 5x⁴
• Negative powers: x^(-3) → -3x^(-4)
• Fractional powers: x^(1/2) → (1/2)x^(-1/2)

Constant rule: If f(x) = c (a constant), then f'(x) = 0

Constant multiple rule: If f(x) = c·g(x), then f'(x) = c·g'(x)

Sum and difference rules: If f(x) = g(x) ± h(x), then f'(x) = g'(x) ± h'(x)

Standard derivatives to memorize:

• d/dx(sin x) = cos x
• d/dx(cos x) = -sin x
• d/dx(e^x) = e^x
• d/dx(ln x) = 1/x
• d/dx(tan x) = sec²x

The product rule

When differentiating the product of two functions, you cannot simply multiply their derivatives. Instead, use:

If y = u(x)·v(x), then dy/dx = u(dv/dx) + v(du/dx)

Alternatively written: (uv)' = u'v + uv'

Method:

1. Identify the two functions being multiplied (u and v)
2. Differentiate each function separately
3. Apply the product rule formula
4. Simplify the result

The product rule extends to three or more functions, though this becomes increasingly complex. For three functions: (uvw)' = u'vw + uv'w + uvw'

The quotient rule

To differentiate a quotient of two functions:

If y = u(x)/v(x), then dy/dx = [v(du/dx) - u(dv/dx)]/v²

Alternatively: (u/v)' = (u'v - uv')/v²

Memory aid: "Low d-high minus high d-low, over the square of what's below"

Method:

1. Identify the numerator (u) and denominator (v)
2. Differentiate each separately
3. Apply the quotient rule formula carefully
4. Simplify, ensuring the denominator is v²

Note: Many quotient problems can alternatively be solved using the product rule by rewriting u/v as u·v^(-1), though the quotient rule is often more efficient.

The chain rule

The chain rule allows differentiation of composite functions — functions within functions.

If y = f(g(x)), then dy/dx = f'(g(x))·g'(x)

Using Leibniz notation: if y = f(u) and u = g(x), then dy/dx = (dy/du)·(du/dx)

Method:

1. Identify the outer function f and inner function g
2. Differentiate the outer function with respect to the inner function
3. Multiply by the derivative of the inner function
4. Simplify

Common applications:

• Powers of functions: d/dx[g(x)]^n = n[g(x)]^(n-1)·g'(x)
• Exponentials: d/dx(e^(g(x))) = e^(g(x))·g'(x)
• Logarithms: d/dx(ln[g(x)]) = g'(x)/g(x)
• Trigonometric: d/dx(sin[g(x)]) = cos[g(x)]·g'(x)

The chain rule often combines with other rules (product, quotient) in complex problems.

Higher-order derivatives

The derivative of a derivative gives the second derivative, denoted f''(x) or d²y/dx².

Geometrically, the second derivative indicates the concavity of the original function:

• f''(x) > 0: function is concave up (cup-shaped)
• f''(x) < 0: function is concave down (cap-shaped)
• f''(x) = 0: possible inflection point

Physically, if s(t) is position, then:

• s'(t) is velocity
• s''(t) is acceleration

Third and higher derivatives follow the same pattern: f'''(x), f⁽⁴⁾(x), etc., or d³y/dx³, d⁴y/dx⁴, etc.

Worked examples

Example 1: Derivative from first principles

Question: Use the limit definition to find the derivative of f(x) = 3x² - 2x.

Solution:

f'(x) = lim[h→0] [f(x + h) - f(x)]/h

First, find f(x + h): f(x + h) = 3(x + h)² - 2(x + h) = 3(x² + 2xh + h²) - 2x - 2h = 3x² + 6xh + 3h² - 2x - 2h

Now substitute: f'(x) = lim[h→0] [(3x² + 6xh + 3h² - 2x - 2h) - (3x² - 2x)]/h = lim[h→0] [6xh + 3h² - 2h]/h = lim[h→0] [h(6x + 3h - 2)]/h = lim[h→0] (6x + 3h - 2)

As h → 0: f'(x) = 6x - 2

Mark scheme notes: 1 mark for correct expansion of f(x + h), 1 mark for correct simplification of numerator, 1 mark for canceling h, 1 mark for correct limit.

Example 2: Combining differentiation rules

Question: Find dy/dx for y = (2x³ + 5)(x² - 3x).

Solution:

This requires the product rule. Let u = 2x³ + 5 and v = x² - 3x.

du/dx = 6x² dv/dx = 2x - 3

Using (uv)' = u'v + uv':

dy/dx = (6x²)(x² - 3x) + (2x³ + 5)(2x - 3) = 6x⁴ - 18x³ + 4x⁴ - 6x³ + 10x - 15 = 10x⁴ - 24x³ + 10x - 15

Alternative method: Expand first, then differentiate: y = 2x⁵ - 6x⁴ + 5x² - 15x dy/dx = 10x⁴ - 24x³ + 10x - 15

Both methods are valid; choose based on efficiency.

Example 3: Chain rule application

Question: A particle's position is given by s(t) = 3sin(2t²) metres, where t is time in seconds. Find the velocity at t = 1 second.

Solution:

Velocity v(t) = ds/dt

This requires the chain rule twice. Let u = 2t², so s = 3sin(u).

ds/du = 3cos(u) du/dt = 4t

By the chain rule: ds/dt = (ds/du)·(du/dt) = 3cos(2t²)·4t = 12t·cos(2t²)

At t = 1: v(1) = 12(1)·cos(2·1²) = 12cos(2) ≈ 12(-0.416) ≈ -4.99 m/s

Mark scheme notes: 1 mark for identifying chain rule needed, 1 mark for correct outer derivative, 1 mark for correct inner derivative, 1 mark for substitution and final answer with units.

Common mistakes and how to avoid them

• Multiplying derivatives in the product rule: Students often write (uv)' = u'v' instead of u'v + uv'. Remember: you must add two terms, each containing one derivative and one original function.

• Sign errors in the quotient rule: The numerator is v(du/dx) - u(dv/dx), not the reverse. The minus sign is crucial. Always write "low d-high minus high d-low" to maintain correct order.

• Forgetting the chain rule: When differentiating sin(3x), the answer is NOT cos(3x). You must multiply by the derivative of the inner function: 3cos(3x). Any function composition requires the chain rule.

• Incorrect power rule application: When differentiating 1/x², rewrite as x^(-2) first, giving -2x^(-3) or -2/x³. Do not attempt to use the power rule on fractions without rewriting.

• Confusing d/dx with f(x): The notation d/dx is an operator meaning "differentiate with respect to x." Write d/dx(x³) = 3x², not d/dx = 3x².

• Mishandling constants: The derivative of 5x³ is 15x², not 5x². Multiply the coefficient by the power. Conversely, d/dx(5) = 0, as constants have zero derivative.

Exam technique for "Differentiation: Definition and Basic Rules"

• Identify which rule(s) to use before starting: Scan the function for products, quotients, or composite functions. Planning prevents errors and saves time. If multiple rules apply, work from the outside in.

• Show clear working for multi-mark questions: Write out u and v when using product or quotient rule. State du/dx and dv/dx separately before substituting. Examiners award method marks even if arithmetic errors occur later.

• Simplify appropriately: Exam mark schemes often require simplified final answers. Combine like terms, factor where obvious, but avoid excessive manipulation that invites arithmetic errors. Leave answers in exact form unless told otherwise.

• Check dimensions and reasonableness: If finding velocity from position, ensure your derivative has correct units (metres per second, not metres). Negative derivatives indicate decreasing functions; verify this makes physical sense in context.

Quick revision summary

The derivative measures instantaneous rate of change and equals the gradient of the tangent line. From first principles: f'(x) = lim[h→0] [f(x+h) - f(x)]/h. Apply the power rule for x^n terms: multiply by the power, reduce power by one. For products use uv' + vu'; for quotients use (vu' - uv')/v². The chain rule handles compositions: differentiate outside, multiply by inside's derivative. Always identify which rule(s) apply before differentiating, show clear working, and simplify final answers.