The Product and Quotient Rules
The derivative of a product is not the product of the derivatives — and the derivative of a quotient is not the quotient of the derivatives. Two dedicated rules handle these combinations: \((fg)' = f'g + fg'\) and \((f/g)' = (f'g - fg')/g^2\).
By the end you'll be able to differentiate products like \(x^2 \sin x\) with the product rule, differentiate quotients with the quotient rule (in the right order), and explain why the tempting shortcut \(f' \cdot g'\) gives the wrong slope.
Predict: does the slope of \(f \cdot g\) equal \(f' \cdot g'\)? Below is the product \(P(x) = x^2 \sin x\), built from \(f(x) = x^2\) and \(g(x) = \sin x\). Slide x and compare the true tangent (green) with the naive \(f'\cdot g'\) guess (grey dashed) — they rarely match.
The true slope comes from the product rule, and the readout splits it into its two ingredients: f′g = 2x·sin x (the change coming from \(f\)) plus fg′ = x²·cos x (the change coming from \(g\)). Hover or focus the green point, or hover the curve, for exact values.
True slope = f′g = 2.99 + fg′ = 0.16 = 3.15 · naive f′·g′ = 0.21 — not the same line.
The product rule, \((fg)' = f'g + fg'\), and the quotient rule, \((f/g)' = (f'g - fg')/g^2\), differentiate products and quotients of functions.
The product rule
When you differentiate a product, you differentiate one factor at a time and add the results:
$$ (f \cdot g)' = f'\,g + f\,g' $$
Each term lets one factor change while the other holds still — \(f'g\) captures the change coming from \(f\), and \(fg'\) the change coming from \(g\). Those are exactly the two colored terms in the demo's readout above, and their sum is the green tangent's slope. The grey dashed line — slope \(f' \cdot g'\) — is what you'd get by wrongly multiplying the derivatives, and you saw it peel away from the curve at almost every \(x\).
If \(f\) and \(g\) are differentiable at \(x\), so are their product and (where \(g(x) \neq 0\)) their quotient, with \((fg)'(x) = f'(x)g(x) + f(x)g'(x)\) and \(\left(\dfrac{f}{g}\right)'(x) = \dfrac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\). Both follow from the limit definition of the derivative — the "More info" block below sketches the proof for products.
Picture revenue as price × quantity, where both drift over time. Revenue's rate of change isn't (price rate) × (quantity rate) — it's (price rate) × quantity + price × (quantity rate): each factor contributes its own change, scaled by the current size of the other. That's the product rule doing real work, and it's why the naive multiply-the-rates guess in the demo lands on the wrong line.
Differentiate \(y = x^2 \sin x\). Let \(f = x^2\) (so \(f' = 2x\), by the power rule) and \(g = \sin x\) (so \(g' = \cos x\), a fact you'll meet properly in the next lesson). Then $$ y' = f'g + fg' = 2x\,\sin x + x^2\cos x. $$ That's the exact green-tangent slope the demo computes at every \(x\).
The quotient rule
Quotients get their own rule:
$$ \left(\frac{f}{g}\right)' = \frac{f'\,g - f\,g'}{g^2} $$
A popular mnemonic is "low d-high minus high d-low, over low-low": denominator times derivative of numerator, minus numerator times derivative of denominator, all over the denominator squared. Unlike the product rule's sum, the numerator here is a difference — and subtraction is not commutative, so the order matters. The numerator's derivative (\(f'g\)) comes first; swap the terms and every answer flips sign.
Differentiate \(y = \dfrac{2x + 1}{x^2}\). Numerator \(f = 2x+1\), so \(f' = 2\); denominator \(g = x^2\), so \(g' = 2x\). Then $$ y' = \frac{2 \cdot x^2 - (2x+1)\cdot 2x}{x^4} = \frac{2x^2 - 4x^2 - 2x}{x^4} = \frac{-2x^2 - 2x}{x^4} = -\frac{2x + 2}{x^3}. $$
Differentiate \(y = \dfrac{3x - 2}{x^2}\). Numerator \(f = 3x - 2\), so \(f' = 3\); denominator \(g = x^2\), so \(g' = 2x\). Fill in the blank: $$ y' = \frac{3 \cdot x^2 - (3x-2)\cdot \_\_\_\_}{x^4} = \; ? $$
Reveal the answer
The blank is \(g' = 2x\): $$ y' = \frac{3x^2 - (3x-2)(2x)}{x^4} = \frac{3x^2 - 6x^2 + 4x}{x^4} = \frac{-3x^2 + 4x}{x^4} = \frac{4 - 3x}{x^3}. $$ Note the numerator kept \(f'g\) first — reversing it would give \(\frac{3x - 4}{x^3}\), wrong by a sign everywhere.
More info — why the product rule works (and where the quotient rule comes from)
Go back to the limit definition of the derivative from earlier in the course. The change in the product is \(f(x+h)g(x+h) - f(x)g(x)\). Add and subtract the mixed term \(f(x+h)g(x)\) and it splits cleanly: $$ f(x+h)\big[g(x+h) - g(x)\big] + g(x)\big[f(x+h) - f(x)\big]. $$ Divide by \(h\) and let \(h \to 0\): the first piece tends to \(f g'\) and the second to \(g f'\) — together, \(f'g + fg'\). The quotient rule then follows by writing \(f/g\) as \(f \cdot g^{-1}\) and differentiating that product (the chain rule, coming later in this section, finishes the job). The OpenStax link in Dive deeper below carries out both proofs in full.
Check your understanding
Differentiate \(y = x^3 \cos x\). (Recall \(\frac{d}{dx}[\cos x] = -\sin x\).)
In the quotient rule for \((f/g)'\), the numerator is a difference of two products. Which statement is right?
Differentiate \(y = \dfrac{x}{x^2 + 1}\).
Differentiate \(y = (x^2 + 3)(2x - 1)\) using the product rule. (You'll need the power and sum rules to differentiate each factor first.)
Recap
- Product rule: \((fg)' = f'g + fg'\) — differentiate one factor at a time and add.
- Quotient rule: \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\) — "low d-high minus high d-low, over low-low".
- \((fg)' \neq f' \cdot g'\) and \((f/g)' \neq f'/g'\) — derivatives don't multiply or divide across products and quotients.
- In the quotient rule the numerator's derivative comes first; reversing the subtraction flips the sign of the answer.
Dive deeper
- Paul's Online Notes — Product and Quotient Rule State both rules and show that (fg)' is not f'g'
- OpenStax Calculus Volume 1 — 3.3 Differentiation Rules Prove the product and quotient rules from the limit definition
Sources
- The Product and Quotient Rules