Unit 2: Differentiation — Definition & Fundamental Properties

Product & Quotient Rules

Differentiating using the product rule and quotient rule.

Advanced25 min lesson4 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Power & Sum Rules

Why the Derivative of a Product Isn't the Product of Derivatives

It's tempting to guess that d/dx[f(x)g(x)] = f'(x)g'(x) — derivatives distributed over sums, so why not products? Before reading on: test that guess on f(x) = x and g(x) = x, where f(x)g(x) = x² has a known derivative, 2x. Does f'(x)g'(x) = 1 · 1 = 1 match?

It doesn't match, so a genuine product rule is needed — and it can be derived from the difference quotient using an add-and-subtract trick. Write [f(x+h)g(x+h) − f(x)g(x)]/h, then insert and immediately remove f(x+h)g(x) in the numerator: f(x+h)g(x+h) − f(x+h)g(x) + f(x+h)g(x) − f(x)g(x). Group into two fractions: f(x+h)·[g(x+h) − g(x)]/h + g(x)·[f(x+h) − f(x)]/h. As h → 0, the first bracket's quotient becomes g'(x), the second becomes f'(x), and f(x+h) → f(x) by continuity (guaranteed by differentiability, from an earlier lesson). The result: f(x)g'(x) + g(x)f'(x).

Definition — Product and Quotient Rules

For differentiable f and g: (fg)' = f'g + fg' (the product rule), and (f/g)' = (f'g − fg')/g², where g(x) ≠ 0 (the quotient rule). The quotient rule can itself be derived from the product rule: writing h = f/g means f = hg, so f' = h'g + hg' by the product rule — solving for h' gives h' = (f' − hg')/g = (f' − (f/g)g')/g, which simplifies to (f'g − fg')/g² after multiplying through by g/g.

\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

Worked Example — Applying the Product Rule

Differentiate f(x) = (x² + 1)(3x − 2). With u = x² + 1, v = 3x − 2: u' = 2x, v' = 3. f'(x) = u'v + uv' = 2x(3x − 2) + (x² + 1)(3) = 6x² − 4x + 3x² + 3 = 9x² − 4x + 3. Check by expanding first: (x² + 1)(3x − 2) = 3x³ − 2x² + 3x − 2, and differentiating directly gives 9x² − 4x + 3 — matching exactly.

Worked Example — Applying the Quotient Rule

Differentiate g(x) = (2x + 1)/(x − 3). With u = 2x + 1, v = x − 3: u' = 2, v' = 1. g'(x) = (u'v − uv')/v² = [2(x − 3) − (2x + 1)(1)]/(x − 3)² = (2x − 6 − 2x − 1)/(x − 3)² = −7/(x − 3)². Check by rewriting first: since 2x + 1 = 2(x − 3) + 7, g(x) = 2 + 7(x − 3)⁻¹, so g'(x) = 7·(−1)(x − 3)⁻² = −7/(x − 3)² — matching exactly.

Derivative Explorer

f(x) =

x^3 - 3*x

Point of tangency

x = a1

Show tangent line

Trace mode

f(x), with tangent line at x = a

Show f''(x)Show f'''(x)

Derivative graph — linked to the graph above; panning or zooming either moves both

What's happening at x = a

At x = 1, f(x) ≈ -2 — the point (1, -2).
The derivative there is f'(1) ≈ 0: the slope of the tangent line, and the instantaneous rate of change of f at this exact point.
That slope is positive, so f is increasing at this point.
The second derivative is positive here, so f is concave up (curving upward) near this point.
Near x = 1, the tangent line y ≈ -2 + 0·(x − 1) is f's best straight-line approximation — the core idea a derivative captures.

Numerical derivative comparison

Forward, backward, and central difference estimates of f'(a) as the step size h shrinks — all three should converge toward the same value.
h	Forward	Backward	Central
1	4	-2	1
0.1	0.31	-0.29	0.01
0.01	0.0301	-0.0299	0.0001
0.001	0.003	-0.003	0

Analysis of f(x)

y-intercept: (0, 0)
x-intercepts: (-1.73, 0), (0, 0), (1.73, 0)
Extrema: local max at (-1, 2); local min at (1, -2)
Inflection points: (0, 0)
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-970, 970]

Analysis of f'(x) — its roots are f's critical points

y-intercept: (0, -3)
x-intercepts: (-1, 0), (1, 0)
Extrema: local min at (0, -3)
Inflection points: none found in view
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-3, 297]

Tip

Verify a product or quotient rule result whenever the original expression can be expanded or rewritten first — two independent methods landing on the same derivative is strong evidence neither has an error.

Common Mistakes

Differentiating (x² + 1)(3x − 2) as though it were f'(x)g'(x) = 2x · 3 = 6x.
There is no shortcut multiplying the two derivatives together — the product rule's f'g + fg' has two full terms, each pairing one function's derivative with the other function unchanged.
Reversing the subtraction order in the quotient rule numerator, writing fg' − f'g instead of f'g − fg'.
Order matters in the quotient rule's numerator — f'g − fg', with the derivative of the numerator function listed first — reversing it flips the sign of the entire result.

Key Takeaways

The product rule, (fg)' = f'g + fg', comes from an add-and-subtract trick applied to the difference quotient of a product.
The quotient rule, (f/g)' = (f'g − fg')/g², can itself be derived from the product rule by writing f = (f/g)·g and solving for (f/g)'.
Expanding a product (or rewriting a quotient) before differentiating, when possible, is a reliable way to check a product- or quotient-rule computation.

Summary

This closes Unit 2: the derivative's definition, plus the power, sum, product, and quotient rules, handle any function built from sums, products, and quotients of powers of x. Unit 3 extends differentiation to functions built by composing one function inside another.

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