The Product Rule and Quotient Rule

By the end of this lesson, you will be able to differentiate products and quotients of functions using the Product and Quotient Rules, know why $(fg)' \neq f'g'$, and apply these rules in complex settings.

The Product Rule

$$(fg)' = f'g + fg'$$

"Derivative of first times second plus first times derivative of second."

Why not $(fg)' = f'g'$? Consider $f = x$ and $g = x$: $(fg)' = (x^2)' = 2x$, but $f'g' = 1 \cdot 1 = 1 \neq 2x$.

The Quotient Rule

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

"Low d-high minus high d-low, over low squared."

Memory device: "Lo dHi minus Hi dLo, over Lo squared"

Example 1 — Product Rule

$h(x) = (3x^2 - 5)(x^4 + 2x)$

$f = 3x^2-5$, $f' = 6x$; $g = x^4+2x$, $g' = 4x^3+2$

$h'(x) = 6x(x^4+2x) + (3x^2-5)(4x^3+2)$

$= 6x^5 + 12x^2 + 12x^5 + 6x^2 - 20x^3 - 10$

$= 18x^5 - 20x^3 + 18x^2 - 10$

Check: could also expand $h(x)$ first then differentiate — same answer.

Example 2 — Quotient Rule

$f(x) = \dfrac{x^2 + 1}{x - 3}$

$\text{num} = x^2+1$, $\text{num}' = 2x$; $\text{den} = x-3$, $\text{den}' = 1$

$f'(x) = \dfrac{2x(x-3) - (x^2+1)(1)}{(x-3)^2} = \dfrac{2x^2-6x-x^2-1}{(x-3)^2} = \dfrac{x^2-6x-1}{(x-3)^2}$

Example 3 — Extending to $e^x$

Using the derivative $\dfrac{d}{dx}[e^x] = e^x$ (proved next lesson):

$f(x) = xe^x$

$f'(x) = e^x + xe^x = e^x(1+x)$

Example 4 — Product of Three Functions

$h = fgk$: write as $h = (fg)k$ and apply product rule twice:

$h' = (fg)'k + (fg)k' = (f'g + fg')k + (fg)k' = f'gk + fg'k + fgk'$

Symmetry: each function gets differentiated once, with the others remaining.

Example 5 — Quotient to Power

$\left(\dfrac{f}{g}\right)^n$: use chain rule (next lesson), but recognize:

$\dfrac{d}{dx}[g^{-1}] = -g^{-2}g'$ (power rule with chain rule); then $\dfrac{d}{dx}[fg^{-1}] = f'g^{-1} + f(-g^{-2}g') = \dfrac{f'g - fg'}{g^2}$ — quotient rule. ✓

Common Mistakes

Mistake 1: $(fg)' = f'g'$ — Always wrong. Use the product rule.