Product Rule (Edexcel A-Level Mathematics): Revision Notes

If you have two functions $u(x)$ and $v(x)$, and you want to differentiate their product with respect to x, the product rule states:

$\frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

In other words, to differentiate the product of two functions:

• Differentiate the first function $u(x)$ and multiply it by the second function $v(x)$ as it is.
• Then, differentiate the second function $v(x)$ and multiply it by the first function $u(x)$ as it is.
• Finally, add these two results together.

Example 1: Differentiate $y = x^2 \cdot \sin(x)$

• Step 1: Identify the functions:
• $u(x) = x^2$
• $v(x) = \sin(x)$
• Step 2: Differentiate each function:
• $u'(x) = 2x$
• $v'(x) = \cos(x)$
• Step 3: Apply the product rule:

$\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

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

$\frac{dy}{dx} = x^2 \cos(x) + 2x \sin(x)$

Example 2: Differentiate $y = e^x \cdot \ln(x)$

• Step 1: Identify the functions:
• $u(x) = e^x$
• $v(x) = \ln(x)$
• Step 2: Differentiate each function:
• $u'(x) = e^x$
• $v'(x) = \frac{1}{x}$
• Step 3: Apply the product rule:

$\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

$\frac{dy}{dx} = e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x$

$\frac{dy}{dx} = \frac{e^x}{x} + e^x \ln(x)$

Example 3: Differentiate $y = (3x^2 + 2x) \cdot \cos(x)$

• Step 1: Identify the functions:
• $u(x) = 3x^2 + 2x$
• $v(x) = \cos(x)$
• Step 2: Differentiate each function:
• $u'(x) = 6x + 2$
• $v'(x) = -\sin(x)$
• Step 3: Apply the product rule:

$\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

$\frac{dy}{dx} = (3x^2 + 2x) \cdot (-\sin(x)) + \cos(x) \cdot (6x + 2)$

$\frac{dy}{dx} = -(3x^2 + 2x)\sin(x) + (6x + 2)\cos(x)$

Example: Differentiate $y = x^2 \cdot e^{3x}$

• Step 1: Identify the functions:
• $u(x) = x^2$
• $v(x) = e^{3x}$
• Step 2: Differentiate each function:
• $u'(x) = 2x$
• $v'(x) = e^{3x} \cdot 3$ (using the chain rule)
• Step 3: Apply the product rule:

$\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)$

$\frac{dy}{dx} = x^2 \cdot 3e^{3x} + e^{3x} \cdot 2x$

$\frac{dy}{dx} = 3x^2 e^{3x} + 2x e^{3x} = e^{3x}(3x^2 + 2x)$

• The product rule is essential for differentiating products of functions, allowing you to break down the differentiation process into manageable steps.
• By differentiating each function separately and then combining the results, you can accurately compute the derivative of more complex expressions.
• Mastery of the product rule, along with other differentiation techniques like the chain rule, is crucial for solving a wide range of problems in calculus and its applications.

Example 1: If $y = x \sin x$, find $\frac{dy}{dx}$.

• Let $u = x$ and $v = \sin x$.
• Then, $u' = 1$ and $v' = \cos x$. Using the product rule:

Example 2: If $y = x^3 (6x^2 + 2)^8$, find $\frac{dy}{dx}$.

• Let $u = x^3$ and $v = (6x^2 + 2)^8$.
• Then, $u' = 3x^2$ and $v' = 8(6x^2 + 2)^7 \times (12x)$. Using the product rule:

Find $\dfrac {dy}{dx}$, simplifying your answer in each case and fully factories

1. $y = 3x^4 e^{2x+3}$

• $u = 3x^4, \quad v = e^{2x+3}$
• $u' = 12x^3, \quad v' = 2e^{2x+3}$

$y = (x+2)(x-3)^3$

Given the function: $y = x^2 \sqrt{3x+1}$

We can set $u = x^2$ and $v = (3x+1)^{\frac{1}{2}}$.

Then, the derivatives are:

$u' = 2x$

Using the product rule:

Combine the terms:

Factoring out common terms:

Simplifying further:

Final simplified form: