1) Introduction: Beyond Sums and Differences - Products and Quotients
In Topic 2, we mastered basic differentiation rules for constants, powers of \( x \), sums, and differences. However, we often encounter functions that are formed by multiplying or dividing other functions. For these situations, the basic rules alone are not enough. We need new rules: the Product Rule and the Quotient Rule.
These rules are essential for differentiating more complex algebraic functions, rational functions (quotients of polynomials), and combinations of different types of functions. Learning these rules will significantly expand your ability to differentiate a wider range of functions.
Why Not Just Differentiate Term by Term for Products and Quotients?
It's a common mistake for beginners to assume that the derivative of a product is just the product of the derivatives (and similarly for quotients). This is generally NOT TRUE! The Product and Quotient Rules provide the correct way to differentiate products and quotients. We need to follow these specific formulas.
2) The Product Rule: Differentiating Products of Functions
The Product Rule tells us how to find the derivative of a function that is formed by multiplying two differentiable functions together.
Rule 3.1: Product Rule
If \( f(x) \) and \( g(x) \) are differentiable functions, then the derivative of their product is given by:
\( \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \)
Using prime notation: \( (f \cdot g)' = f' \cdot g + f \cdot g' \)
In words: The derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. It's often remembered as: "first derivative times second plus first times second derivative."
Examples of the Product Rule
- Example 1: \( y = x^2 \sin(x) \) (Product of a polynomial and a trigonometric function - for illustration, we'll only differentiate \(x^2 \cdot x^3\) here first)
Let \( f(x) = x^2 \) and \( g(x) = x^3 \). We want to find the derivative of \( y = f(x)g(x) = x^2 \cdot x^3 = x^5 \). (We know the derivative should be \( 5x^4 \) using the Power Rule. Let's check using the Product Rule).
- \( f'(x) = \frac{d}{dx}[x^2] = 2x \)
- \( g'(x) = \frac{d}{dx}[x^3] = 3x^2 \)
- Example 2: \( h(x) = (3x^2 + 5x)(x^3 - 2) \) (Product of two polynomials)
Let \( f(x) = 3x^2 + 5x \) and \( g(x) = x^3 - 2 \).
- \( f'(x) = \frac{d}{dx}[3x^2 + 5x] = 6x + 5 \) (using Sum and Power Rules)
- \( g'(x) = \frac{d}{dx}[x^3 - 2] = 3x^2 - 0 = 3x^2 \) (using Power and Constant Rules)
Now, let's simplify this expression:
\( = (6x^4 - 12x + 5x^3 - 10) + (9x^4 + 15x^3) \) \( = 6x^4 + 5x^3 - 12x - 10 + 9x^4 + 15x^3 \) \( = (6x^4 + 9x^4) + (5x^3 + 15x^3) - 12x - 10 \) \( = 15x^4 + 20x^3 - 12x - 10 \) So, \( h'(x) = 15x^4 + 20x^3 - 12x - 10 \). - Example 3: \( y = \sqrt{x} (x^2 + 1) \) (Product involving a radical)
Let \( f(x) = \sqrt{x} = x^{1/2} \) and \( g(x) = x^2 + 1 \).
- \( f'(x) = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \) (using Power Rule)
- \( g'(x) = \frac{d}{dx}[x^2 + 1] = 2x \) (using Sum and Power Rules)
We can simplify this:
\( = \frac{x^2 + 1}{2\sqrt{x}} + 2x\sqrt{x} \)To combine into a single fraction, use a common denominator of \( 2\sqrt{x} \):
\( = \frac{x^2 + 1}{2\sqrt{x}} + \frac{2x\sqrt{x} \cdot 2\sqrt{x}}{2\sqrt{x}} = \frac{x^2 + 1}{2\sqrt{x}} + \frac{4x(\sqrt{x})^2}{2\sqrt{x}} = \frac{x^2 + 1}{2\sqrt{x}} + \frac{4x \cdot x}{2\sqrt{x}} \) \( = \frac{x^2 + 1 + 4x^2}{2\sqrt{x}} = \frac{5x^2 + 1}{2\sqrt{x}} \) So, \( \frac{dy}{dx} = \frac{5x^2 + 1}{2\sqrt{x}} \).
3) The Quotient Rule: Differentiating Quotients of Functions
The Quotient Rule is used to find the derivative of a function that is a ratio (quotient) of two differentiable functions.
Rule 3.2: Quotient Rule
If \( f(x) \) and \( g(x) \) are differentiable functions, with \( g(x) \neq 0 \), then the derivative of their quotient is given by:
\( \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2} \)
Using prime notation: \( \left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2} \)
In words: The derivative of a quotient is "derivative of the top times the bottom minus the top times the derivative of the bottom, all over the bottom squared." A helpful mnemonic is often used: "low dee high minus high dee low, over low-low" (where "high" refers to the numerator \( f(x) \) and "low" to the denominator \( g(x) \)). Make sure to get the subtraction order correct in the numerator.
Examples of the Quotient Rule
- Example 1: \( y = \frac{x^3}{x^2 + 1} \) (Quotient of two polynomials)
Let \( f(x) = x^3 \) and \( g(x) = x^2 + 1 \).
- \( f'(x) = \frac{d}{dx}[x^3] = 3x^2 \) (using Power Rule)
- \( g'(x) = \frac{d}{dx}[x^2 + 1] = 2x \) (using Sum and Power Rules)
Simplify the numerator:
\( = \frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} = \frac{x^4 + 3x^2}{(x^2 + 1)^2} \) So, \( \frac{dy}{dx} = \frac{x^4 + 3x^2}{(x^2 + 1)^2} \). - Example 2: \( h(x) = \frac{\sqrt{x}}{x + 3} \) (Quotient involving a radical)
Let \( f(x) = \sqrt{x} = x^{1/2} \) and \( g(x) = x + 3 \).
- \( f'(x) = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \) (using Power Rule)
- \( g'(x) = \frac{d}{dx}[x + 3] = 1 \) (using Sum, Power, and Constant Rules)
Simplify the numerator - let's multiply numerator and denominator of the main fraction by \( 2\sqrt{x} \) to clear the fraction in the numerator:
\( = \frac{\left[\left(\frac{1}{2\sqrt{x}}\right)(x + 3) - \sqrt{x}\right] \cdot 2\sqrt{x}}{(x + 3)^2 \cdot 2\sqrt{x}} = \frac{(x + 3) - 2x}{(x + 3)^2 \cdot 2\sqrt{x}} = \frac{3 - x}{2\sqrt{x}(x + 3)^2} \) So, \( h'(x) = \frac{3 - x}{2\sqrt{x}(x + 3)^2} \).
4) Combining Product and Quotient Rules: More Complex Derivatives
Example 1: Combining Product and Sum Rules
Find the derivative of \( y = x^3(2x + 5) \).
We can use the Product Rule with \( f(x) = x^3 \) and \( g(x) = 2x + 5 \).
- \( f'(x) = 3x^2 \)
- \( g'(x) = 2 \)
Simplify:
\( = 6x^3 + 15x^2 + 2x^3 = 8x^3 + 15x^2 \) So, \( \frac{dy}{dx} = 8x^3 + 15x^2 \). (Note: For this particular example, we *could* have also just expanded \( y = x^3(2x + 5) = 2x^4 + 5x^3 \) and then differentiated term by term using basic rules - try it and see you get the same answer!)Example 2: Quotient Rule and Constant Multiple
Find the derivative of \( f(x) = \frac{4x^2}{x - 1} \).
We use the Quotient Rule with \( f(x) = 4x^2 \) (numerator - top) and \( g(x) = x - 1 \) (denominator - bottom).
- \( f'(x) = \frac{d}{dx}[4x^2] = 8x \)
- \( g'(x) = \frac{d}{dx}[x - 1] = 1 \)
Simplify the numerator:
\( = \frac{8x^2 - 8x - 4x^2}{(x - 1)^2} = \frac{4x^2 - 8x}{(x - 1)^2} \) So, \( f'(x) = \frac{4x^2 - 8x}{(x - 1)^2} \). We could factor out \( 4x \) from the numerator if needed: \( f'(x) = \frac{4x(x - 2)}{(x - 1)^2} \).5) Note Box: Common Mistakes and Things to Watch Out For
Avoiding Common Errors with Product and Quotient Rules
Here are some frequent mistakes to avoid when using the Product and Quotient Rules:
- Incorrectly Assuming Simple Product/Quotient of Derivatives: Remember, \( (f \cdot g)' \neq f' \cdot g' \) and \( (\frac{f}{g})' \neq \frac{f'}{g'} \) in general. You must use the specific formulas for the Product and Quotient Rules.
- Order of Terms in Quotient Rule Numerator: The Quotient Rule numerator is \( f'g - fg' \), NOT \( fg' - f'g \). The subtraction order matters! Remember "low dee high minus high dee low".
- Forgetting to Square the Denominator in Quotient Rule: The denominator of the Quotient Rule is \( g(x)^2 \). Don't forget to square the original denominator function.
- Algebraic Simplification Errors: After applying the rules, be careful with algebraic simplification (expanding, combining like terms, factoring). Errors in simplification are common.
- Unnecessary Quotient Rule: Sometimes, you can rewrite a quotient as a product using negative exponents and then use the Product Rule instead (or simpler rules). For example, \( \frac{x^2}{x^3} = x^{-1} \) - easier to differentiate \( x^{-1} \) directly than using Quotient Rule. Or \( \frac{x^2}{\sqrt{x}} = x^{2 - 1/2} = x^{3/2} \). Simplify *before* differentiating if possible!
6) Practice Questions - Product and Quotient Rules
Test your proficiency with the Product and Quotient Rules.
Fundamental Practice Questions
Instructions: Find the derivative \( f'(x) \) or \( \frac{dy}{dx} \) for each function using the Product Rule, Quotient Rule, or basic rules as needed.
- Q1. \( y = x^4(x^2 + 3) \) (Product Rule)
- Q2. \( f(x) = (2x - 1)(x^3 + x) \) (Product Rule)
- Q3. \( g(x) = \frac{x^5}{x^2 + 2} \) (Quotient Rule)
- Q4. \( h(x) = \frac{x - 3}{x + 3} \) (Quotient Rule)
- Q5. \( y = \frac{3x^2}{x^3 - 1} \) (Quotient Rule)
- Q6. \( f(x) = \sqrt{x}(x^2 - 4x) \) (Product Rule, rewrite \( \sqrt{x} = x^{1/2} \))
- Q7. \( y = (x^2 + 2)(x^{-2} - 2) \) (Product Rule, rewrite \( x^{-2} = \frac{1}{x^2} \))
- Q8. \( g(x) = \frac{5}{x^3 + 2x - 1} \) (Quotient Rule - treat 5 as \( f(x) \) and denominator as \( g(x) \))
- Q9. \( h(x) = \frac{x^2 - 1}{x^2 + 1} \) (Quotient Rule)
- Q10. \( y = (x^3 + 2x - 3)\sqrt{x} \) (Product Rule)
- Q11. \( f(x) = \frac{x}{\sqrt{x} + 1} \) (Quotient Rule, rewrite \( \sqrt{x} = x^{1/2} \))
- Q12. \( y = \frac{x^4 - 5x^2 + 2}{3} \) (Simplify first - divide each term by 3, then differentiate using basic rules. Or use Quotient Rule with constant denominator - compare methods!)
Challenging Practice Questions
Instructions: These problems may involve more complex algebra, combinations of rules, or conceptual thinking.
- Q1. Find the equation of the tangent line to the curve \( y = \frac{x^2}{x + 2} \) at the point \( x = 2 \). (You'll need the point and the derivative using the Quotient Rule).
- Q2. Find the derivative of \( F(x) = \frac{(x^2 + 1)(x - 1)}{x^3} \). (You can use Product Rule in numerator and then Quotient Rule for the whole fraction, OR simplify numerator first, OR divide term by term after expanding numerator - explore different approaches).
- Q3. For what value(s) of \( x \) does the function \( g(x) = \frac{x^2}{x - 2} \) have a horizontal tangent line? (Set \( g'(x) = 0 \) and solve for \( x \)).
- Q4. If \( f(x) \) and \( g(x) \) are differentiable functions, and you know that \( f(3) = 2 \), \( f'(3) = -1 \), \( g(3) = 4 \), and \( g'(3) = 3 \), find the value of \( \left(\frac{f}{g}\right)'(3) \) using the Quotient Rule.
- Q5. Consider the function \( h(x) = \frac{x}{\sqrt{x} + 1} \). Find \( h'(x) \). Then find \( h'(4) \) and interpret what \( h'(4) \) represents about the rate of change of \( h(x) \) at \( x = 4 \). What are the units of \( h'(4) \) if \( x \) represents time in seconds and \( h(x) \) represents distance in meters?
7) Summary & Cheat Sheet - Product and Quotient Rules
Quick review of the Product and Quotient Rules.
7.1) Product Rule
To differentiate \( y = f(x)g(x) \): \( \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \) or \( (f \cdot g)' = f' \cdot g + f \cdot g' \)
"Derivative of first times second plus first times derivative of second."
7.2) Quotient Rule
To differentiate \( y = \frac{f(x)}{g(x)} \): \( \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2} \) or \( \left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2} \)
"Low dee high minus high dee low, over low-low" (Derivative of bottom times top minus top times derivative of bottom, over bottom squared).
7.3) Key Points to Remember
- Product and Quotient Rules are essential for products and quotients of functions.
- Follow the formulas precisely – order and signs matter, especially in the Quotient Rule.
- Algebraic simplification is often necessary after applying these rules.
- Sometimes, simplifying *before* differentiating (using algebra or exponent rules) can make differentiation easier.
Excellent work! You've now expanded your differentiation toolkit with the Product and Quotient Rules, enabling you to tackle a much broader range of functions. Next, we'll explore the powerful Chain Rule!