AOS3 Topic 7: Product Rule
The product rule is a general rule for differentiation problems where one function is multiplied by another function. The derivative of the product of two differentiable functions is equal to the sum of the first function multiplied by the derivative of the second, and the second function multiplied by the derivative of the first function. The functions may be exponential, logarithmic, or of other types.
Product Rule Formula
Let \( F(x) = f(x) \cdot g(x) \). If \( f'(x) \) and \( g'(x) \) exist, then
\[ F'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x) \]
For example, consider \( F(x) = (x^2 + 3x)(4x + 5) \). Then \( F \) is the product of two functions \( f \) and \( g \), where \( f(x) = x^2 + 3x \) and \( g(x) = 4x + 5 \). The product rule gives:
\[ F'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x) \]
\[ = (x^2 + 3x) \cdot 4 + (4x + 5) \cdot (2x + 3) \]
\[ = 4x^2 + 12x + 8x^2 + 22x + 15 \]
\[ = 12x^2 + 34x + 15 \]
This could also have been found by multiplying \( x^2 + 3x \) by \( 4x + 5 \) and then differentiating.
The Product Rule (Function Notation)
Let \( F(x) = f(x) \cdot g(x) \). If \( f'(x) \) and \( g'(x) \) exist, then
\[ F'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x) \]
Proof
By the definition of the derivative of \( F \), we have
\[ F'(x) = \lim_{h \to 0} \frac{F(x + h) - F(x)}{h} = \lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x) g(x)}{h} \]
Adding and subtracting \( f(x + h) g(x) \):
\[ F'(x) = \lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x) g(x) + f(x + h) g(x) - f(x + h) g(x)}{h} \]
\[ = \lim_{h \to 0} \left[ f(x + h) \cdot \frac{g(x + h) - g(x)}{h} + g(x) \cdot \frac{f(x + h) - f(x)}{h} \right] \]
Since \( f \) and \( g \) are differentiable, we obtain
\[ F'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x) \]
We can state the product rule in Leibniz notation and give a geometric interpretation.
The Product Rule (Leibniz Notation)
If \( y = uv \), where \( u \) and \( v \) are functions of \( x \), then
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
In the following figure, the white region represents \( y = uv \) and the shaded region \( \delta y \), as explained below.
\(\delta v\) \(u \delta v\) \( \delta u \delta v\)
\(v\) \(uv\) \(v \delta u\)
\(u\) \(\delta u\)
\(\delta y = (u + \delta u)(v + \delta v) - uv\)
\(= uv + v \delta u + u \delta v + \delta u \delta v - uv\)
\(= v \delta u + u \delta v + \delta u \delta v\)
\(\therefore \frac{\delta y}{\delta x} = v \frac{\delta u}{\delta x} + u \frac{\delta v}{\delta x} + \frac{\delta u}{\delta x} \frac{\delta v}{\delta x} \delta x\)
In the limit, as \(\delta x \to 0\), we have
\(\frac{\delta u}{\delta x} = \frac{du}{dx}, \quad \frac{\delta v}{\delta x} = \frac{dv}{dx}, \quad \text{and} \quad \frac{\delta y}{\delta x} = \frac{dy}{dx}\)
Therefore
\(\frac{dy}{dx} = v \frac{du}{dx} + u \frac{dv}{dx}\)
Example 1
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = 12x^3(3x − 5)^3 + (3x − 5)^4 \cdot 3x^2 = (3x − 5)^3 \left(12x^3 + 3x^2(3x − 5)\right) = (3x − 5)^3 \left(12x^3 + 9x^3 − 15x^2\right) = (3x − 5)^3 \left(21x^3 − 15x^2\right) = 3x^2(7x − 5)(3x − 5)^3 \]
Example 2
\[ F'(x) = x^{-3} \cdot 60x(10x^2 − 5)^2 + (10x^2 − 5)^3 \cdot (−3x^{-4}) = (10x^2 − 5)^2 \left(60x^{-2} + (10x^2 − 5) \cdot (−3x^{-4})\right) = (10x^2 − 5)^2 \left(\frac{60x^2 − 30x^2 + 15}{x^4}\right) = \frac{(10x^2 − 5)^2 (30x^2 + 15)}{x^4} \]
Example 3
\[ \frac{dy}{dx} = e^x \sqrt{x - 1} + \frac{1}{2} e^x (x - 1)^{-\frac{1}{2}} \]
\[ = e^x \sqrt{x - 1} + \frac{e^x}{2 \sqrt{x - 1}} \]
\[ = \frac{2e^x (x - 1) + e^x}{2 \sqrt{x - 1}} \]
\[ = \frac{2x e^x - e^x}{2 \sqrt{x - 1}} \]
Example 4
- \( 2x^2 \sin(2x) \)
- \( e^{2x} \sin(2x + 1) \)
\[ \frac{dy}{dx} = 2x^2 \cdot \frac{d}{dx}[\sin(2x)] + \sin(2x) \cdot \frac{d}{dx}[2x^2] \]
\[ \frac{dy}{dx} = 2x^2 \cdot 2 \cos(2x) + \sin(2x) \cdot 4x \]
\[ \frac{dy}{dx} = 4x \sin(2x) + 4x^2 \cos(2x) \]
\[ \frac{dy}{dx} = e^{2x} \cdot \frac{d}{dx}[\sin(2x + 1)] + \sin(2x + 1) \cdot \frac{d}{dx}[e^{2x}] \]
\[ \frac{dy}{dx} = e^{2x} \cdot 2 \cos(2x + 1) + \sin(2x + 1) \cdot 2e^{2x} \]
\[ \frac{dy}{dx} = 2e^{2x} \cos(2x + 1) + 2e^{2x} \sin(2x + 1) \]