Product Rule Notes

Chapter 3: Differentiation Rules - The Product Rule

Introduction

• The Product Rule is a method for finding the derivative of a product of two functions.
• An initial, incorrect guess was that the derivative of a product is the product of the derivatives like Leibniz's initial thought.
• For example, if $f(x) = x^2$ and $g(x) = x^3$, then $f'(x) = 2x$ and $g'(x) = 3x^2$. But $f(x)g(x) = x^5$, so $(f(x)g(x))' = 5x^4$, which is not equal to $(2x)(3x^2) = 6x^3$.

Discovery of the Product Rule

• Consider two positive differentiable functions, $u$ and $v$.
• The product $uv$ can be interpreted as the area of a rectangle.
• If $u$ changes by $\Delta u$ and $v$ changes by $\Delta v$, the new value of the product is $(u + \Delta u)(v + \Delta v)$, which can be interpreted as the area of a larger rectangle.
• The change in the area of the rectangle is:
  $(u + \Delta u)(v + \Delta v) - uv = u\Delta v + v\Delta u + \Delta u \Delta v$
• Dividing by $\Delta x$:
  $\frac{(u + \Delta u)(v + \Delta v) - uv}{\Delta x} = u\frac{\Delta v}{\Delta x} + v\frac{\Delta u}{\Delta x} + \frac{\Delta u}{\Delta x} \Delta v$

Leibniz Notation and the Definition of a Derivative

• The definition of a derivative in Leibniz notation is:
  $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$

The Product Rule

• If $u$ and $v$ are both differentiable, then:
  $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
• In prime notation:
  $(uv)' = u'v + uv'$

Proof of the Product Rule

• Starting with:
  $\frac{\Delta (uv)}{\Delta x} = u\frac{\Delta v}{\Delta x} + v\frac{\Delta u}{\Delta x} + \frac{\Delta u}{\Delta x} \Delta v$
• Taking the limit as $\Delta x$ approaches 0:
  $\lim<em>{\Delta x \to 0} \frac{\Delta (uv)}{\Delta x} = \lim</em>{\Delta x \to 0} \left(u\frac{\Delta v}{\Delta x} + v\frac{\Delta u}{\Delta x} + \frac{\Delta u}{\Delta x} \Delta v\right)$
• Since $u$ is differentiable, it is also continuous, so $\lim_{\Delta x \to 0} \Delta u = 0$
• Thus,
  $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} + 0 = u\frac{dv}{dx} + v\frac{du}{dx}$
• The Product Rule holds for all differentiable functions $u$ and $v$, whether they are positive or negative.
• In words: The derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Example 1

• Differentiate the function $f(x) = x^2e^x$
• Solution:
  • Using the Product Rule:
    $f'(x) = x^2 \frac{d}{dx}(e^x) + e^x \frac{d}{dx}(x^2) = x^2e^x + e^x(2x) = x^2e^x + 2xe^x$
  • If $f(x) = x^2e^x$, find $f'''(x)$
    • First derivative: $f'(x) = x^2e^x + 2xe^x$
    • Second derivative: $f''(x) = (x^2e^x + 2xe^x) + (2xe^x + 2e^x) = x^2e^x + 4xe^x + 2e^x$
    • Third derivative: $f'''(x) = (x^2e^x + 4xe^x + 2e^x) + (2xe^x + 4e^x + 0) = x^2e^x + 6xe^x + 6e^x$
• Each successive differentiation adds another term with $e^x$
• The graph of $f(x)$ is increasing when $f'(x)$ is positive and decreasing when $f'(x)$ is negative.

Example 2

• Differentiate the function $y = \sqrt{x}(x^2 + 1)$
• Solution 1:
  • Using the Product Rule:
    $\frac{dy}{dx} = \sqrt{x}\frac{d}{dx}(x^2 + 1) + (x^2 + 1)\frac{d}{dx}(\sqrt{x}) = \sqrt{x}(2x) + (x^2 + 1)\left(\frac{1}{2\sqrt{x}}\right) = 2x^{3/2} + \frac{x^2 + 1}{2\sqrt{x}}$
• Solution 2:
  • Rewriting the function using laws of exponents:
    $y = x^{1/2}(x^2 + 1) = x^{5/2} + x^{1/2}$
  • Differentiating directly:
    $\frac{dy}{dx} = \frac{5}{2}x^{3/2} + \frac{1}{2}x^{-1/2}$
  • This answer is equivalent to the answer in Solution 1.

Constants and Variables

• In mathematics, letters near the beginning of the alphabet (e.g., $a, b, c$) typically represent constants, while letters near the end of the alphabet (e.g., $x, y, z$) represent variables.

Example 3

• If $g(x) = f(x) \sin(x)$, where $f(\frac{\pi}{6}) = 4$ and $f'(\frac{\pi}{6}) = -2$, find $g'(\frac{\pi}{6})$.
• Solution:
  • Applying the Product Rule:
    $g'(x) = f(x) \frac{d}{dx}(\sin(x)) + \sin(x) \frac{d}{dx}(f(x)) = f(x)\cos(x) + \sin(x)f'(x)$
  • Evaluating at $x = \frac{\pi}{6}$:
    $g'\left(\frac{\pi}{6}\right) = f\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right)f'\left(\frac{\pi}{6}\right) = 4\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)(-2) = 2\sqrt{3} - 1$

Simplification

• Sometimes it is easier to simplify a product of functions before differentiating than to use the Product Rule such as in Example 2.
• In Example 1, the Product Rule is the only possible method because we cannot simplify $x^2e^x$ by using exponent rules or any other technique.