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

Differentiate with respect to x: \(x^3(3x − 5)^4\)

Solution:

Let \(y = x^3(3x − 5)^4\). Let \(u = x^3\) and \(v = (3x − 5)^4\).

Then \(\frac{du}{dx} = 3x^2\) and \(\frac{dv}{dx} = 12(3x − 5)^3\), using the chain rule.

The product rule gives:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = 12x^3(3x − 5)^3 + (3x − 5)^4 \cdot 3x^2 = (3x − 5)^3 \left(12x^3 + 3x^2(3x − 5)\right) = (3x − 5)^3 \left(12x^3 + 9x^3 − 15x^2\right) = (3x − 5)^3 \left(21x^3 − 15x^2\right) = 3x^2(7x − 5)(3x − 5)^3 \]

Example 2

For \(F : \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}\), \(F(x) = x^{-3}(10x^2 − 5)^3\), find \(F'(x)\).

Solution:

Let \(f(x) = x^{-3}\) and \(g(x) = (10x^2 − 5)^3\).

Then \(f'(x) = −3x^{-4}\) and \(g'(x) = 60x(10x^2 − 5)^2\) using the chain rule.

\[ F'(x) = x^{-3} \cdot 60x(10x^2 − 5)^2 + (10x^2 − 5)^3 \cdot (−3x^{-4}) = (10x^2 − 5)^2 \left(60x^{-2} + (10x^2 − 5) \cdot (−3x^{-4})\right) = (10x^2 − 5)^2 \left(\frac{60x^2 − 30x^2 + 15}{x^4}\right) = \frac{(10x^2 − 5)^2 (30x^2 + 15)}{x^4} \]

Example 3

Differentiate the following with respect to \( x \): \( e^x \sqrt{x - 1} \).

Solution:

Use the product rule and the chain rule.

Let \( y = e^x \sqrt{x - 1} \). Then

\[ \frac{dy}{dx} = e^x \sqrt{x - 1} + \frac{1}{2} e^x (x - 1)^{-\frac{1}{2}} \]

\[ = e^x \sqrt{x - 1} + \frac{e^x}{2 \sqrt{x - 1}} \]

\[ = \frac{2e^x (x - 1) + e^x}{2 \sqrt{x - 1}} \]

\[ = \frac{2x e^x - e^x}{2 \sqrt{x - 1}} \]

Example 4

Question: Find the derivative of each of the following with respect to \( x \):

  • \( 2x^2 \sin(2x) \)
  • \( e^{2x} \sin(2x + 1) \)

Solution:

a) Let \( y = 2x^2 \sin(2x) \). Applying the product rule:

\[ \frac{dy}{dx} = 2x^2 \cdot \frac{d}{dx}[\sin(2x)] + \sin(2x) \cdot \frac{d}{dx}[2x^2] \]

\[ \frac{dy}{dx} = 2x^2 \cdot 2 \cos(2x) + \sin(2x) \cdot 4x \]

\[ \frac{dy}{dx} = 4x \sin(2x) + 4x^2 \cos(2x) \]

b) Let \( y = e^{2x} \sin(2x + 1) \). Applying the product rule:

\[ \frac{dy}{dx} = e^{2x} \cdot \frac{d}{dx}[\sin(2x + 1)] + \sin(2x + 1) \cdot \frac{d}{dx}[e^{2x}] \]

\[ \frac{dy}{dx} = e^{2x} \cdot 2 \cos(2x + 1) + \sin(2x + 1) \cdot 2e^{2x} \]

\[ \frac{dy}{dx} = 2e^{2x} \cos(2x + 1) + 2e^{2x} \sin(2x + 1) \]

Exercise &&1&& (&&1&& Question)

Differentiate with respect to \( x \): \((2x^2 + 1)(5x^3 + 16)\)

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Exercise &&2&& (&&1&& Question)

Differentiate the following with respect to \( x \):

\( e^x (2x^2 + 1) \)

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Exercise &&3&& (&&1&& Question)

Find the derivative with respect to \( x \):

\( \cos(4x) \sin(2x) \)

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