Edexcel IAL - Pure Maths 3- 4.2 Differentiation using the product rule, quotient rule and chain rule- Study notes - New syllabus
Edexcel IAL – Pure Maths 3- 4.2 Differentiation using the product rule, quotient rule and chain rule -Study notes- New syllabus
Edexcel IAL – Pure Maths 3- 4.2 Differentiation using the product rule, quotient rule and chain rule -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 4.2 Differentiation using the product rule, quotient rule and chain rule
The Product Rule (Differentiation)
The product rule is used when a function is written as the product of two functions of \( x \).
Statement of the Product Rule
If:
\( y = u(x)\,v(x) \)
then:
\( \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \)
This means:
Differentiate one function at a time, keeping the other unchanged.
When to Use the Product Rule
- When functions are multiplied together
- When expanding first would be complicated
- Common with polynomial × trig or exponential functions
Method
- Identify \( u \) and \( v \)
- Differentiate \( u \) and \( v \)
- Substitute into the product rule formula
- Simplify the final expression
Example
Differentiate:
\( y = x\sin x \)
▶️ Answer / Explanation
Let:
\( u = x,\quad v = \sin x \)
\( \dfrac{du}{dx} = 1,\quad \dfrac{dv}{dx} = \cos x \)
Apply the product rule:
\( \dfrac{dy}{dx} = x\cos x + \sin x \)
Example
Differentiate:
\( y = 2x^4 \sin x \)
▶️ Answer / Explanation
Let:
\( u = 2x^4,\quad v = \sin x \)
\( \dfrac{du}{dx} = 8x^3,\quad \dfrac{dv}{dx} = \cos x \)
Apply the product rule:
\( \dfrac{dy}{dx} = 2x^4\cos x + 8x^3\sin x \)
Example
Differentiate:
\( y = x^2 e^x \)
▶️ Answer / Explanation
Let:
\( u = x^2,\quad v = e^x \)
\( \dfrac{du}{dx} = 2x,\quad \dfrac{dv}{dx} = e^x \)
Apply the product rule:
\( \dfrac{dy}{dx} = x^2 e^x + 2x e^x \)
Factorise:
\( \dfrac{dy}{dx} = e^x(x^2 + 2x) \)
The Quotient Rule (Differentiation)
The quotient rule is used when a function is written as one function divided by another.
Statement of the Quotient Rule
If:
\( y = \dfrac{u(x)}{v(x)} \)
then:
\( \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} – u\dfrac{dv}{dx}}{v^2} \)
This can be remembered as:
Bottom × derivative of top − Top × derivative of bottom, all over bottom squared.
When to Use the Quotient Rule
- When one function is divided by another
- When simplifying first is not convenient
- Common with exponential or trigonometric numerators
Method
- Identify the numerator \( u \)
- Identify the denominator \( v \)
- Differentiate \( u \) and \( v \)
- Substitute carefully into the quotient rule formula
- Simplify the final expression
Example
Differentiate:
\( y = \dfrac{x^2}{x+1} \)
▶️ Answer / Explanation
Let:
\( u = x^2,\quad v = x+1 \)
\( \dfrac{du}{dx} = 2x,\quad \dfrac{dv}{dx} = 1 \)
Apply the quotient rule:
\( \dfrac{dy}{dx} = \dfrac{(x+1)(2x) – x^2(1)}{(x+1)^2} \)
\( = \dfrac{x^2 + 2x}{(x+1)^2} \)
Example
Differentiate:
\( y = \dfrac{e^x}{x} \)
▶️ Answer / Explanation
Let:
\( u = e^x,\quad v = x \)
\( \dfrac{du}{dx} = e^x,\quad \dfrac{dv}{dx} = 1 \)
Apply the quotient rule:
\( \dfrac{dy}{dx} = \dfrac{x e^x – e^x}{x^2} \)
Factorise:
\( \dfrac{dy}{dx} = \dfrac{e^x(x – 1)}{x^2} \)
Example
Differentiate:
\( y = \dfrac{\sin x}{x^2} \)
▶️ Answer / Explanation
Let:
\( u = \sin x,\quad v = x^2 \)
\( \dfrac{du}{dx} = \cos x,\quad \dfrac{dv}{dx} = 2x \)
Apply the quotient rule:
\( \dfrac{dy}{dx} = \dfrac{x^2\cos x – \sin x(2x)}{x^4} \)
\( = \dfrac{x\cos x – 2\sin x}{x^3} \)
The Chain Rule (Differentiation)
The chain rule is used when a function is composed of another function, that is, when one function is inside another.
Statement of the Chain Rule
If:
\( y = f(g(x)) \)
then:
\( \dfrac{dy}{dx} = \dfrac{dy}{dg} \cdot \dfrac{dg}{dx} \)
This means:
Differentiate the outer function first, then multiply by the derivative of the inner function.
When to Use the Chain Rule
- When the argument is not just \( x \)
- Functions involving powers, trigonometric or exponential functions of another expression
- Expressions such as \( \cos x^2 \), \( e^{3x} \), \( \tan(2x) \)
Method
- Identify the inner function
- Differentiate the outer function
- Multiply by the derivative of the inner function
- Write the final answer clearly
Example
Differentiate:
\( y = \cos(2x) \)
▶️ Answer / Explanation
Let:
\( u = 2x \)
\( y = \cos u \)
\( \dfrac{dy}{du} = -\sin u,\quad \dfrac{du}{dx} = 2 \)
\( \dfrac{dy}{dx} = -2\sin(2x) \)
Example
Differentiate:
\( y = e^{3x} \)
▶️ Answer / Explanation
Let:
\( u = 3x \)
\( y = e^u \)
\( \dfrac{dy}{du} = e^u,\quad \dfrac{du}{dx} = 3 \)
\( \dfrac{dy}{dx} = 3e^{3x} \)
Example
Differentiate:
\( y = \tan^2(2x) \)
▶️ Answer / Explanation
Rewrite:
\( y = (\tan(2x))^2 \)
Let:
\( u = \tan(2x) \)
\( \dfrac{dy}{du} = 2u \)
Now differentiate \( u \):
\( \dfrac{du}{dx} = 2\sec^2(2x) \)
Apply the chain rule:
\( \dfrac{dy}{dx} = 4\tan(2x)\sec^2(2x) \)