Calculus: Derivative Rules — Product, Quotient and Chain Rule

TL;DR

When you have complex functions that are products, quotients, or compositions of simpler functions, you'll need specific rules to find their derivatives. The Product Rule handles multiplication, the Quotient Rule tackles division, and the Chain Rule applies when one function is inside another. Mastering these three rules lets you differentiate a huge variety of functions.

1. The Mental Model

Think of these rules as specialized tools in your calculus toolbox. Just like you wouldn't use a screwdriver to hammer a nail, you wouldn't use the basic power rule for a function like $e^x \sin(x)$. Each rule is designed for a specific way functions interact.

2. The Core Material

When functions get combined through multiplication, division, or nesting, their derivatives aren't as simple as just applying basic rules to each piece. That's where the Product, Quotient, and Chain Rules come in.

The Product Rule

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This rule helps you find the derivative of a function that's the product of two other functions, say $f(x) = u(x)v(x)$. It's not just the product of their derivatives!

The Product Rule states: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

A common way to remember this is "the derivative of the first times the second, plus the first times the derivative of the second."

Example: Find the derivative of $f(x) = x^2 \sin(x)$.
Here, let $u(x) = x^2$ and $v(x) = \sin(x)$.
Then $u'(x) = 2x$ and $v'(x) = \cos(x)$.
Applying the Product Rule: $f'(x) = (2x)(\sin(x)) + (x^2)(\cos(x)) = 2x\sin(x) + x^2\cos(x)$.

The Quotient Rule

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This rule is for functions that are ratios (quotients) of two other functions, like $f(x) = \frac{u(x)}{v(x)}$. There's a catchy phrase to help remember it.

The Quotient Rule states: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

To remember it, think "low d-high minus high d-low, all over low squared." ("low" is $v(x)$, "high" is $u(x)$, "d-high" is $u'(x)$, "d-low" is $v'(x)$).

Example: Find the derivative of $f(x) = \frac{e^x}{x}$.
Here, let $u(x) = e^x$ and $v(x) = x$.
Then $u'(x) = e^x$ and $v'(x) = 1$.
Applying the Quotient Rule: $f'(x) = \frac{(e^x)(x) - (e^x)(1)}{x^2} = \frac{xe^x - e^x}{x^2}$.

The Chain Rule

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This is arguably the most powerful and frequently used of the three. It's for composite functions, where one function is "inside" another, like $f(x) = g(h(x))$.

The Chain Rule states: If $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

Think of it as taking the derivative of the "outer" function first, leaving the "inner" function untouched, and then multiplying by the derivative of the "inner" function.

Example: Find the derivative of $f(x) = \sin(x^3)$.
Here, the outer function is $\sin(\text{something})$ and the inner function is $x^3$.
Let $g(u) = \sin(u)$ and $h(x) = x^3$.
Then $g'(u) = \cos(u)$ and $h'(x) = 3x^2$.
Applying the Chain Rule: $f'(x) = \cos(x^3) \cdot 3x^2 = 3x^2\cos(x^3)$.

Here's a diagram showing when to apply each rule:

graph TD
    A["Function to Differentiate"] --> B{Is it a sum/difference?}
    B -- Yes --> C["Differentiate term by term"]
    B -- No --> D{Is it a product (u*v)?}
    D -- Yes --> E["Apply Product Rule: u'v + uv'"]
    D -- No --> F{Is it a quotient (u/v)?}
    F -- Yes --> G["Apply Quotient Rule: (u'v - uv') / v^2"]
    F -- No --> H{Is it a composite (f(g(x)))?}
    H -- Yes --> I["Apply Chain Rule: f'(g(x)) * g'(x)"]
    H -- No --> J["Apply basic rules (power, exponential, trig)"]

3. Worked Example

Let's find the derivative of $f(x) = (x^2+1)^3 e^{5x}$. This function combines the Product Rule and the Chain Rule.

1. Identify the main structure: This is a product of two functions: $(x^2+1)^3$ and $e^{5x}$. So, we'll start with the Product Rule.
   Let $u(x) = (x^2+1)^3$ and $v(x) = e^{5x}$.
   The Product Rule says $f'(x) = u'(x)v(x) + u(x)v'(x)$.

2. Find $u'(x)$: For $u(x) = (x^2+1)^3$, we need the Chain Rule.
   Outer function: $(\text{something})^3$. Its derivative is $3(\text{something})^2$.
   Inner function: $(x^2+1)$. Its derivative is $2x$.
   So, $u'(x) = 3(x^2+1)^2 \cdot (2x) = 6x(x^2+1)^2$.

3. Find $v'(x)$: For $v(x) = e^{5x}$, we also need the Chain Rule.
   Outer function: $e^{\text{something}}$. Its derivative is $e^{\text{something}}$.
   Inner function: $5x$. Its derivative is $5$.
   So, $v'(x) = e^{5x} \cdot 5 = 5e^{5x}$.

4. Put it all together with the Product Rule:
   $f'(x) = [6x(x^2+1)^2] \cdot [e^{5x}] + [(x^2+1)^3] \cdot [5e^{5x}]$

5. Simplify (optional but good practice):
   Notice that both terms have common factors: $(x^2+1)^2$ and $e^{5x}$.
   $f'(x) = e^{5x}(x^2+1)^2 [6x + 5(x^2+1)]$
   $f'(x) = e^{5x}(x^2+1)^2 [6x + 5x^2 + 5]$
   $f'(x) = e^{5x}(x^2+1)^2 (5x^2 + 6x + 5)$

4. Key Takeaways

• The Product Rule applies to functions multiplied together: $u'v + uv'$.
• The Quotient Rule applies to functions divided: $\frac{u'v - uv'}{v^2}$.
• The Chain Rule applies to composite functions: $f'(g(x)) \cdot g'(x)$.
• Always identify the main structure of the function first to decide which rule to apply initially.
• You'll often need to use these rules in combination, as seen in the worked example.
• Don't confuse the derivative of a product with the product of derivatives – they are different.

Common Mistakes to Avoid:
- Product Rule: Forgetting the "+" sign or the specific order, thinking it's simply $u'v'$.
- Quotient Rule: Mixing up the order of subtraction in the numerator ($uv' - u'v$ instead of $u'v - uv'$), or forgetting to square the denominator.
- Chain Rule: Forgetting to multiply by the derivative of the inner function, or taking the derivative of both outer and inner functions at the same time.
- Applying a rule when it's not needed (e.g., differentiating $3x^2$ with the Product Rule instead of the Power Rule).

5. Now Try It

Take 15 minutes to find the derivative of the following function:

$g(x) = \frac{\cos(4x)}{x^3+2}$

Think about the main rule needed first, then identify any nested functions that require the Chain Rule. Break it down into steps: identify $u$ and $v$, find $u'$ and $v'$ (using the Chain Rule where necessary), then apply the Quotient Rule.

Success looks like a correct final derivative, $g'(x) = \frac{-4\sin(4x)(x^3+2) - 3x^2\cos(4x)}{(x^3+2)^2}$, showing your intermediate steps clearly.