Differentiation

CIE A-Level Mathematics· 6 min read

1. Core Differentiation Rules★★☆☆☆⏱ 20 min

For combinations of functions, we use three core rules to find derivatives, building on basic differentiation of standard functions.

📘 Definition

Chain Rule

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

Rule for differentiating composite functions $y = f (g (x))$ : differentiate the outer function, multiply by derivative of the inner function.

Product Rule: For $y = uv$ , $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
Quotient Rule: For $y = \frac{u}{v}$ , $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$

📐 Worked Example

Differentiate $y = e^{2 x} sin (3 x)$ with respect to $x$

1
Identify that $y$ is a product of two functions: $u = e^{2 x}$ , $v = sin (3 x)$
2
Differentiate $u$ and $v$ using the chain rule:
3
$\frac{du}{dx} = 2e^{2x}, \quad \frac{dv}{dx} = 3\cos(3x)$
4
Apply the product rule:
5
$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = e^{2x}(3\cos 3x) + \sin(3x)(2e^{2x})$
6
Factor out the common term $e^{2 x}$ to simplify:
7
$\frac{dy}{dx} = e^{2x}(3\cos 3x + 2\sin 3x)$

Exam tip:

Always simplify and factor your final derivative; exam markers award full marks only for simplified expressions.

2. Implicit Differentiation★★★☆☆⏱ 25 min

Many relations between $x$ and $y$ cannot be rearranged to the explicit form $y = f (x)$ . We use implicit differentiation to find $\frac{d y}{d x}$ for these cases.

📘 Definition

Implicit Differentiation

Differentiate every term on both sides of the equation with respect to $x$ , applying the chain rule to all terms containing $y$ .

📐 Worked Example

Find $\frac{d y}{d x}$ at the point $(3, 4)$ for the circle $x^{2} + y^{2} = 25$

1
Differentiate each term with respect to $x$ :
2
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$
3
Apply the chain rule to $y^{2}$ : $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$
4
$2x + 2y \frac{dy}{dx} = 0$
5
Rearrange to isolate $\frac{d y}{d x}$ :
6
$\frac{dy}{dx} = -\frac{x}{y}$
7
Substitute $x = 3$ , $y = 4$ :
8
$\frac{dy}{dx} = -\frac{3}{4}$

3. Parametric Differentiation★★★★☆⏱ 25 min

When a curve is defined by parametric equations $x = x (t)$ , $y = y (t)$ , where $t$ is the parameter, we rearrange the chain rule to find $\frac{d y}{d x}$ .

📘 Definition

Parametric Differentiation

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Find derivatives of $x$ and $y$ with respect to $t$ , then divide $\frac{d y}{d t}$ by $\frac{d x}{d t}$ to get $\frac{d y}{d x}$ .

📐 Worked Example

Find $\frac{d y}{d x}$ for the parametric curve $x = 2 cos t$ , $y = 3 sin t$ , $0 < t < \frac{π}{2}$

1
Differentiate $x$ and $y$ with respect to $t$ :
2
$\frac{dx}{dt} = -2\sin t, \quad \frac{dy}{dt} = 3\cos t$
3
Apply the parametric differentiation formula:
4
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t}{-2\sin t} = -\frac{3}{2}\cot t$

Exam tip:

For the second derivative $\frac{d ^{2} y}{d x ^{2}}$ , remember that $\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{d x / d t}$ , do not just differentiate $\frac{d y}{d x}$ directly with respect to $t$ .

4. Applications: Rates of Change★★★☆☆⏱ 20 min

We use the chain rule to relate the rates of change of two quantities that both change with time. This is a common exam application of differentiation.

📐 Worked Example

The radius of a spherical balloon increases at $0.5 cm s^{- 1}$ . Find the rate of increase of volume when radius is $10 cm$ .

1
Volume of a sphere is $V = \frac{4}{3} π r^{3}$ . We know $\frac{d r}{d t} = 0.5$ , need $\frac{d V}{d t}$ .
2
Apply the chain rule:
3
$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$
4
Differentiate $V$ with respect to $r$ :
5
$\frac{dV}{dr} = 4\pi r^2$
6
Substitute $r = 10$ and $\frac{d r}{d t} = 0.5$ :
7
$\frac{dV}{dt} = 4\pi (10)^2 (0.5) = 200\pi \text{ cm}^3\text{s}^{-1}$

5. Common Pitfalls

Wrong move:

Forgetting to multiply by the derivative of the inner function when using the chain rule

Why:

You only differentiate the outer function and miss the inner derivative factor

Correct move:

Always explicitly identify inner and outer functions, do not skip writing the inner derivative step

Wrong move:

Mixing up the order of terms in the quotient rule

Why:

Most people incorrectly reverse the numerator terms

Correct move:

Remember the mnemonic: 'top derivative times bottom minus bottom derivative times top, all over bottom squared'

Wrong move:

Forgetting the $\frac{d y}{d x}$ factor when differentiating $y$ -terms in implicit differentiation

Why:

You accidentally treat $y$ as a constant instead of a function of $x$

Correct move:

Every time you differentiate a term containing $y$ , add a $\frac{d y}{d x}$ factor from the chain rule

Wrong move:

Calculating the second derivative for parametric curves by just differentiating $\frac{d y}{d x}$ with respect to $t$

Why:

You forget $\frac{d y}{d x}$ is still a function of $t$ , not $x$

Correct move:

After differentiating $\frac{d y}{d x}$ with respect to $t$ , divide the result by $\frac{d x}{d t}$ to get $\frac{d ^{2} y}{d x ^{2}}$

Wrong move:

Using mismatched units in rates of change problems

Why:

You forget to convert all quantities to consistent units before starting calculations

Correct move:

Convert all lengths, times and volumes to matching units at the start of the problem

6. Quick Reference Cheatsheet

Rule	Formula
Chain Rule	$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$
Product Rule	$\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}$
Quotient Rule	$\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$
Implicit Differentiation	Differentiate all terms, add $\frac{d y}{d x}$ for $y$ -terms
Parametric First Derivative	$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
Parametric Second Derivative	$\frac{d ^{2} y}{d x ^{2}} = \frac{d / d t ( d y / d x )}{d x / d t}$
Rates of Change	$\frac{d A}{d t} = \frac{d A}{d x} \cdot \frac{d x}{d t}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 3
Implicit differentiation gradient calculation
2022 · 3
Parametric differentiation in rates problem
2021 · 3
Chain rule for exponential-trigonometric product

Going deeper

syllabusOfficial CIE 9709 Syllabus

What's Next

Differentiation is the foundation for almost all remaining calculus topics in CIE P3, including further integration techniques, solving differential equations, and analyzing complex curves. Questions on Paper 3 regularly combine differentiation with other topics like inverse trigonometric functions and exponential growth, so mastery of these techniques is critical for a high grade. Next, you will build on these core differentiation skills to find tangents, normals, stationary points, and solve optimisation problems, before moving on to integration, the inverse operation of differentiation that makes up a large portion of the P3 exam.