Download original notes
PAGE 1

3.4 The Product and the Quotient Rules

We know \[ \frac{d}{dx} [f(x) \pm g(x)] = f'(x) \pm g'(x) \] Sum/difference rules

but \[ \frac{d}{dx} [f(x) g(x)] \neq f'(x) g'(x) \]

The Product Rule takes care of the \( \frac{d}{dx} [f(x) g(x)] \)

\[ \frac{d}{dx} [f(x) g(x)] = f(x) g'(x) + f'(x) g(x) \]

the order is irrelevant

\[ \frac{d}{dx} [g(x) f(x)] = g(x) f'(x) + s'(x) f(x) \]

another notation: \[ (fg)'(x) = f'(x) g(x) + f(x) g'(x) \]

PAGE 2

example

\( y = (1 + x^2)(2 - x^2) \)

let's try this in two ways

1) Product Rule

\( y = \underbrace{(1 + x^2)}_{f} \underbrace{(2 - x^2)}_{g} \)

\( y' = \underbrace{(1 + x^2)}_{f} \underbrace{\frac{d}{dx}(2 - x^2)}_{g'} + \underbrace{(2 - x^2)}_{g} \underbrace{\frac{d}{dx}(1 + x^2)}_{f'} \)

\( = (1 + x^2)(-2x) + (2 - x^2)(2x) \)

\( = -2x - 2x^3 + 4x - 2x^3 = -4x^3 + 2x \)

2) multiply y out first

\( y = 2 - x^2 + 2x^2 - x^4 = 2 + x^2 - x^4 \)

deriv is simple: \( y' = +2x - 4x^3 \)

PAGE 3

Example: Second Derivative with Product Rule

Given the function:

\[ y = x^2 e^x \]

Find \( y'' \).

Starting with the first derivative using the product rule:

\[ y' = (x^2) \frac{d}{dx}(e^x) + (e^x) \frac{d}{dx}(x^2) \]\[ = (x^2)(e^x) + (e^x)(2x) \]\[ = (e^x)(x^2 + 2x) \]

To find the second derivative \( y'' \), we apply the product rule again to the result of \( y' \), where we can let \( f = e^x \) and \( g = x^2 + 2x \):

\[ y'' = (e^x) \frac{d}{dx}(x^2 + 2x) + (x^2 + 2x) \frac{d}{dx}(e^x) \]\[ = (e^x)(2x + 2) + (x^2 + 2x)(e^x) \]\[ = 2xe^x + 2e^x + x^2e^x + 2xe^x \]\[ = x^2e^x + 4xe^x + 2e^x \]
PAGE 4

Now the Quotient Rule

The general formula for the quotient rule is:

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}, \quad g(x) \neq 0 \]

Note the sign: order matters!

A common mnemonic for this is "Low d-High minus High d-Low over Low-Low":

\[ \frac{d}{dx} \left[ \frac{\text{high}}{\text{low}} \right] = \frac{(\text{low}) \cdot (d\text{-high}) - (\text{high}) \cdot (d\text{-low})}{(\text{low})(\text{low})} \]

Example

Differentiate the following function:

\[ y = \frac{6x - 1}{2x - 4} \]

Applying the quotient rule:

\[ y' = \frac{(2x - 4) \frac{d}{dx}(6x - 1) - (6x - 1) \frac{d}{dx}(2x - 4)}{(2x - 4)(2x - 4)} \]\[ = \frac{(2x - 4)(6) - (6x - 1)(2)}{(2x - 4)^2} = \frac{12x - 24 - 12x + 2}{(2x - 4)^2} \]\[ = \frac{-22}{(2x - 4)^2} \]
PAGE 5

Example: Combining Quotient and Product Rules

Consider the function:

\[ y = \frac{xe^x}{x+2} \]

Start with the quotient rule. Note that the numerator \( xe^x \) is a product of two functions of \( x \), so it will also require the product rule during the differentiation process.

\[ y' = \frac{(x+2) \frac{d}{dx}(xe^x) - (xe^x) \frac{d}{dx}(x+2)}{(x+2)^2} \]

Applying the product rule to the term \( \frac{d}{dx}(xe^x) \):

\[ = \frac{(x+2) \left[ (x) \cdot \frac{d}{dx}(e^x) + (e^x) \frac{d}{dx}(x) \right] - (xe^x)(1)}{(x+2)^2} \]
\[ = \frac{(x+2)(xe^x + e^x) - xe^x}{(x+2)^2} \]

Expanding the numerator:

\[ = \frac{x^2e^x + xe^x + 2xe^x + 2e^x - xe^x}{(x+2)^2} \]

Simplifying the final expression:

\[ = \frac{x^2e^x + 2xe^x + 2e^x}{(x+2)^2} \]
PAGE 6

Alternative Strategies for Differentiation

Just like with the product rule, sometimes using the quotient rule directly is not the easiest way.

Example 1: Constant Numerator

For example, consider \( y = \frac{8}{x} \).

Using Quotient Rule

\[ y' = \frac{(x) \frac{d}{dx}(8) - (8) \frac{d}{dx}(x)}{(x)^2} \]\[ = \frac{(x)(0) - (8)}{(x)^2} = -\frac{8}{x^2} \]

Using Power Rule (Easier)

\[ y = 8x^{-1} \]\[ y' = 8(-1 \cdot x^{-2}) = -8x^{-2} = -\frac{8}{x^2} \]

Example 2: Simplifying Fractions

Another example: \( y = \frac{2-x^3}{x^4} \). By splitting the fraction first, the quotient rule is not needed.

\[ y = \frac{2}{x^4} - \frac{x^3}{x^4} = 2x^{-4} - x^{-1} \]

Now differentiate using the power rule:

\[ y' = -8x^{-5} + x^{-2} = -\frac{8}{x^5} + \frac{1}{x^2} \]
PAGE 7

but we have no choice if

\[ y = \frac{x^4}{2-x^3} \neq \frac{x^4}{2} - \frac{x^4}{x^3} \]

quotient rule is the only way

\[ y' = \frac{(2-x^3) \frac{d}{dx}(x^4) - (x^4) \frac{d}{dx}(2-x^3)}{(2-x^3)^2} \]
\[ = \frac{(2-x^3)(4x^3) - (x^4)(-3x^2)}{(2-x^3)^2} \]
\[ = \frac{8x^3 - 4x^6 + 3x^6}{(2-x^3)^2} = \frac{8x^3 - x^6}{(2-x^3)^2} \]
PAGE 8
\[ \frac{d}{dx} [f(x)g(x)] = f(x)g'(x) + g(x)f'(x) \]
Optional

why?

\[ \frac{d}{dx} [f(x)g(x)] = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \]

let's think of \( f(x)g(x) \) as area of a rectangle with length \( f(x) \) and width \( g(x) \), \( f(x+h)g(x+h) \) as a bigger rectangle w/ sides \( f(x+h) \) and \( g(x+h) \)

Geometric proof of product rule using nested rectangles with dimensions f(x), g(x) and f(x+h), g(x+h).

the difference in area between the two rectangles is \( f(x+h)g(x+h) - f(x)g(x) \), which is the numerator of the difference quotient

PAGE 9

Derivation of the Product Rule

We see from the picture that is also the L-shaped area, which can be expressed as:

\[ [f(x+h) - f(x)] g(x) + [g(x+h) - g(x)] f(x) + [f(x+h) - f(x)][g(x+h) - g(x)] \]

area of rectangle mark ①

②

③

Replace numerator of limit with the above:

\[ \frac{d}{dx} [f(x)g(x)] = \lim_{h \to 0} \frac{[f(x+h) - f(x)] g(x) + [g(x+h) - g(x)] f(x) + [f(x+h) - f(x)][g(x+h) - g(x)]}{h} \]

Rewrite:

\[ = \lim_{h \to 0} \left[ \frac{[f(x+h) - f(x)] g(x)}{h} + \frac{[g(x+h) - g(x)] f(x)}{h} + \frac{[f(x+h) - f(x)][g(x+h) - g(x)]}{h} \right] \]
PAGE 10
\[ = \lim_{h \to 0} \frac{[f(x+h) - f(x)]}{h} \cdot g(x) + \lim_{h \to 0} \frac{[g(x+h) - g(x)]}{h} \cdot f(x) + \lim_{h \to 0} \frac{[f(x+h) - f(x)]}{h} \cdot [g(x+h) - g(x)] \]

\( f'(x) \)

\( g'(x) \)

goes to 0 as \( h \to 0 \)

\[ = f'(x) g(x) + g'(x) f(x) \quad \text{product rule.} \]

Quotient rule is much more complicated, but basic idea is similar.