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A Level Mathematics Syllabus Statement

Differentiation

Syllabus Content

Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions

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Furthermore

Examples

1. Chain Rule:
The chain rule is used when differentiating composite functions. If we have a function \( y = f(u) \) and \( u = g(x) \), then the derivative of \( y \) with respect to \( x \) is given by:

$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$

Example: Differentiate \( y = \sin(3x^2) \) with respect to \( x \).

Let \( u = 3x^2 \). Then, \( \frac{du}{dx} = 6x \) and \( \frac{dy}{du} = \cos(u) \). Using the chain rule:

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

2. Product Rule:
The product rule is used when differentiating the product of two functions. If \( y = u \cdot v \) where both \( u \) and \( v \) are functions of \( x \), then the derivative of \( y \) with respect to \( x \) is:

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

Example: Differentiate \( y = x^2 \cdot \ln(x) \) with respect to \( x \).

Using the product rule:

$$ \frac{dy}{dx} = 2x \cdot \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x $$

3. Quotient Rule:
The quotient rule is used when differentiating the quotient of two functions. If \( y = \frac{u}{v} \) where both \( u \) and \( v \) are functions of \( x \) and \( v \neq 0 \), then the derivative of \( y \) with respect to \( x \) is:

$$ \frac{dy}{dx} = \frac{u' \cdot v - u \cdot v'}{v^2} $$

Example: Differentiate \( y = \frac{x^2}{\sin(x)} \) with respect to \( x \).

Using the quotient rule:

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

Here's a 'deep fake' video featuring the images of Arnold Schwarzenegger, Ice Spice and Mr Beast explaining the chain rule for differentiation. While the humans might be fake the maths is real and correct.

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