The power rule alone is not enough when functions are multiplied or divided. You cannot simply differentiate each piece separately — you need special rules for products and quotients.
🎯 In this section you will learn
- The Product Rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
- The Quotient Rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
- When to use each rule (and when not to)
📌 The Product Rule
When you have a product of two functions, the derivative is NOT simply the product of the derivatives. Instead:
$$ \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) $$
💡 Memory Trick
"First times derivative of the second, plus second times derivative of the first."
Find $\frac{d}{dx}(x^2 \cdot \sin x)$.
①
Identify $f$ and $g$
$f(x) = x^2$, $g(x) = \sin x$
②
Find $f'$ and $g'$
$f'(x) = 2x$, $g'(x) = \cos x$
③
Apply Product Rule
$f'g + fg' = (2x)(\sin x) + (x^2)(\cos x)$
④
Simplify
$\frac{d}{dx}(x^2 \sin x) = 2x\sin x + x^2\cos x$
Find $\frac{d}{dx}(x e^x)$.
- $f(x) = x$, $g(x) = e^x$
- $f'(x) = 1$, $g'(x) = e^x$
- $\frac{d}{dx}(x e^x) = (1)(e^x) + (x)(e^x) = e^x + x e^x = e^x(1 + x)$
For products of three functions: $(fgh)' = f'gh + fg'h + fgh'$
- Example: $\frac{d}{dx}(x \cdot e^x \cdot \ln x)$
- $= (1)(e^x)(\ln x) + (x)(e^x)(\ln x) + (x)(e^x)(\frac{1}{x})$
- $= e^x\ln x + x e^x\ln x + e^x$
📌 The Quotient Rule
When you have a fraction (quotient) of two functions:
$$ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $$
💡 Memory Trick
"Low d-high minus high d-low, over low squared."
(low = denominator, high = numerator)
Find $\frac{d}{dx}\left(\frac{x^2}{\sin x}\right)$.
①
Identify $f$ and $g$
$f(x) = x^2$, $g(x) = \sin x$
②
Find $f'$ and $g'$
$f'(x) = 2x$, $g'(x) = \cos x$
③
Apply Quotient Rule
$\frac{f'g - fg'}{g^2} = \frac{(2x)(\sin x) - (x^2)(\cos x)}{\sin^2 x}$
④
Simplify
$\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{2x\sin x - x^2\cos x}{\sin^2 x}$
Find $\frac{d}{dx}\left(\frac{e^x}{x}\right)$.
- $f(x) = e^x$, $g(x) = x$
- $f'(x) = e^x$, $g'(x) = 1$
- $\frac{d}{dx}\left(\frac{e^x}{x}\right) = \frac{(e^x)(x) - (e^x)(1)}{x^2} = \frac{x e^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2}$
Sometimes rewriting can be easier! For $\frac{d}{dx}\left(\frac{x^2 + 1}{x}\right)$, you could use quotient rule OR rewrite:
- Rewrite: $\frac{x^2 + 1}{x} = x + x^{-1}$
- Then $\frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$
- This is much simpler than using the quotient rule!
💡 Tip
Before using the quotient rule, check if you can rewrite the expression as a sum of simpler terms.
⚠️ Common Mistakes
- Product Rule order: It's $f'g + fg'$, not $f'g'$ or $f'g - fg'$.
- Quotient Rule order: It's $\frac{f'g - fg'}{g^2}$ — the minus sign is critical! Many students write $fg' - f'g$ incorrectly.
- Forgetting to square the denominator in the quotient rule.
- Using product rule when functions are multiplied by constants — that's just constant multiple rule, not product rule.
🔍 Key Takeaways
- Product Rule: $\frac{d}{dx}[fg] = f'g + fg'$
- Quotient Rule: $\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}$
- Memory for Quotient Rule: "Low d-high minus high d-low, over low squared"
- When possible, simplify before differentiating (e.g., split fractions into sums).
- Both rules work with any differentiable functions — polynomials, trig, exponentials, logs.
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