Product and Quotient Rule for MATH 122
Exam Relevance for MATH 122
Tested directly on every exam. Often combined with other derivative rules in multi-step problems.
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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