Product and Quotient Rule for MATH 139
Exam Relevance for MATH 139
Used constantly. Memorize both rules and practice combining with chain rule. Common source of errors.
This skill appears on:
The Product Rule
Problem: How do you find the derivative of two functions multiplied together?
Common Mistake: $(fg)' \neq f' \cdot g'$ — you can't just multiply the derivatives!
The Formula
If $h(x) = f(x) \cdot g(x)$, then:
$$h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$$
Memory trick: "First times derivative of second, plus second times derivative of first."
Or think: $(fg)' = f'g + fg'$
Problem: Find $\frac{d}{dx}[x^2 \cdot e^x]$
Step 1: Identify the parts:
- $f(x) = x^2$ → $f'(x) = 2x$
- $g(x) = e^x$ → $g'(x) = e^x$
Step 2: Apply product rule: $$\frac{d}{dx}[x^2 \cdot e^x] = f'g + fg'$$ $$= 2x \cdot e^x + x^2 \cdot e^x$$
Step 3: Factor out common terms: $$= e^x(2x + x^2) = \boxed{e^x(x^2 + 2x)}$$
or equivalently: $\boxed{xe^x(x + 2)}$
Problem: Find $\frac{d}{dx}[x \sin x]$
Step 1: Identify the parts:
- $f(x) = x$ → $f'(x) = 1$
- $g(x) = \sin x$ → $g'(x) = \cos x$
Step 2: Apply product rule: $$(x \sin x)' = f'g + fg' = 1 \cdot \sin x + x \cdot \cos x$$
$$\boxed{\sin x + x\cos x}$$
The Quotient Rule
Problem: How do you find the derivative of one function divided by another?
The Formula
If $h(x) = \frac{f(x)}{g(x)}$, then:
$$h'(x) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$$
Memory tricks:
- "Lo d-Hi minus Hi d-Lo, over Lo-Lo" (lo = denominator, hi = numerator)
- "Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared"
Warning: The order matters! It's $f'g - fg'$, not the other way around.
Problem: Find $\frac{d}{dx}\left[\frac{x^2}{x+1}\right]$
Step 1: Identify the parts:
- $f(x) = x^2$ (top) → $f'(x) = 2x$
- $g(x) = x+1$ (bottom) → $g'(x) = 1$
Step 2: Apply quotient rule: $$\frac{d}{dx}\left[\frac{x^2}{x+1}\right] = \frac{f'g - fg'}{g^2}$$ $$= \frac{2x(x+1) - x^2(1)}{(x+1)^2}$$
Step 3: Simplify the numerator: $$= \frac{2x^2 + 2x - x^2}{(x+1)^2} = \boxed{\frac{x^2 + 2x}{(x+1)^2}}$$
Or factored: $\frac{x(x+2)}{(x+1)^2}$
Problem: Find $\frac{d}{dx}[\tan x]$ using $\tan x = \frac{\sin x}{\cos x}$
Step 1: Identify:
- $f(x) = \sin x$ → $f'(x) = \cos x$
- $g(x) = \cos x$ → $g'(x) = -\sin x$
Step 2: Apply quotient rule: $$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$ $$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$
Step 3: Use Pythagorean identity ($\cos^2 x + \sin^2 x = 1$): $$= \frac{1}{\cos^2 x} = \boxed{\sec^2 x}$$
When to Use Which Rule
| Expression | Rule to Use |
|---|---|
| $f(x) \cdot g(x)$ | Product Rule |
| $\frac{f(x)}{g(x)}$ | Quotient Rule |
| $c \cdot f(x)$ | Constant Multiple (just $c \cdot f'(x)$) |
| $\frac{1}{g(x)}$ | Quotient OR Power Rule: $(g^{-1})' = -g^{-2} \cdot g'$ |
Problem: Find $\frac{d}{dx}\left[\frac{3}{x^2}\right]$
Method 1: Quotient Rule $$\frac{d}{dx}\left[\frac{3}{x^2}\right] = \frac{0 \cdot x^2 - 3 \cdot 2x}{x^4} = \frac{-6x}{x^4} = -\frac{6}{x^3}$$
Method 2: Rewrite and use Power Rule (often easier!) $$\frac{3}{x^2} = 3x^{-2}$$ $$\frac{d}{dx}[3x^{-2}] = 3(-2)x^{-3} = -6x^{-3} = -\frac{6}{x^3}$$
$$\boxed{-\frac{6}{x^3}}$$
Tip: When the numerator is a constant, rewriting is usually faster than using the quotient rule!
Common Mistakes and Misunderstandings
❌ Mistake: Thinking you can just multiply derivatives
Wrong: $(fg)' = f' \cdot g'$
Why it's wrong: The product rule exists precisely because derivatives don't distribute over multiplication.
Example: Let $f(x) = x^2$ and $g(x) = x^3$, so $f(x) \cdot g(x) = x^5$.
- Correct way (product rule): $(x^2 \cdot x^3)' = 2x \cdot x^3 + x^2 \cdot 3x^2 = 2x^4 + 3x^4 = 5x^4$ ✓
- Wrong way (multiplying derivatives): $f' \cdot g' = 2x \cdot 3x^2 = 6x^3$ ✗
We can verify: $(x^5)' = 5x^4$, which matches the product rule, not the wrong method!
Correct: $(fg)' = f'g + fg'$
Product Rule
The derivative of a product of two functions equals the first times the derivative of the second, plus the second times the derivative of the first
Variables:
- $f$:
- first function
- $g$:
- second function
- $f'$:
- derivative of f
- $g'$:
- derivative of g
Quotient Rule
The derivative of a quotient. Remember: derivative of top times bottom MINUS top times derivative of bottom, all over bottom squared
Variables:
- $f$:
- numerator function (top)
- $g$:
- denominator function (bottom)
- $f'$:
- derivative of numerator
- $g'$:
- derivative of denominator
- $g^2$:
- denominator squared
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