14.4 Product and Quotient Rules

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What is a Product? What is a Quotient?

Before diving in, let's clarify two words you'll see constantly:

• A product means two things multiplied together. For example, $x^2 \times (x^5 + 1)$ is a product.
• A quotient means one thing divided by another. For example, $\dfrac{x^3 + 1}{2x - 3}$ is a quotient.

When you need to differentiate (find the derivative of) these types of expressions, you cannot simply differentiate each part separately. You need special rules — the Product Rule and the Quotient Rule.

Part 1: The Product Rule

What is it?

The Product Rule is a formula that lets you differentiate the product of two functions.

If you have: $y = u \cdot v$

where both $u$ and $v$ are functions of $x$ (meaning they both contain $x$), then:

$\boxed{\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}}$

In plain English:

How to Label u and v

When you see a product, simply:

1. Call the first function $u$
2. Call the second function $v$
3. Find $\dfrac{du}{dx}$ (differentiate $u$)
4. Find $\dfrac{dv}{dx}$ (differentiate $v$)
5. Plug everything into the formula

Worked Example 1 — Basic Product Rule

Differentiate $y = x^2(x^5 + 1)$

Step 1 — Label the parts: $u = x^2 \qquad v = x^5 + 1$

Step 2 — Find the derivatives of each part: $\frac{du}{dx} = 2x \qquad \frac{dv}{dx} = 5x^4$

Step 3 — Apply the Product Rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ $= (x^2)(5x^4) + (x^5 + 1)(2x)$ $= 5x^6 + 2x^6 + 2x$ $= 7x^6 + 2x$

Worked Example 2 — Product Rule with a Square Root

Find the derivative of $y = (5x + 1)\sqrt{6x - 1}$

First, rewrite the square root as a power (this makes it easier to differentiate): $y = (5x + 1)(6x - 1)^{\frac{1}{2}}$

Step 1 — Label the parts: $u = 5x + 1 \qquad v = (6x-1)^{\frac{1}{2}}$

Step 2 — Differentiate each part (note: to differentiate $v$, you need the Chain Rule — multiply the power down and reduce the power by 1, then multiply by the derivative of the inside): $\frac{du}{dx} = 5$ $\frac{dv}{dx} = \frac{1}{2}(6x - 1)^{-\frac{1}{2}} \times 6 = \frac{3}{(6x-1)^{\frac{1}{2}}} = \frac{3}{\sqrt{6x-1}}$

Step 3 — Apply the Product Rule: $\frac{dy}{dx} = (5x + 1) \cdot \frac{3}{\sqrt{6x-1}} + \sqrt{6x-1} \cdot 5$

$= \frac{3(5x+1)}{\sqrt{6x-1}} + 5\sqrt{6x-1}$

Step 4 — Simplify (put everything over a common denominator of $\sqrt{6x-1}$):

$= \frac{3(5x+1) + 5(6x-1)}{\sqrt{6x-1}}$

$= \frac{15x + 3 + 30x - 5}{\sqrt{6x-1}}$

$= \frac{45x - 2}{\sqrt{6x-1}}$