Differentiation

Q: How is a function determined to be strictly increasing on an interval?

A function is increasing if its first derivative is positive: \( \frac{dy}{dx} > 0 \).

Mathematics (9709) · Cambridge International A Level · 6 min read · Updated: 15 Nov 2025

Pure Mathematics 3: Comprehensive Study Notes on Differentiation

Welcome to the Differentiation chapter for Pure Mathematics 3! Differentiation is not just about finding the gradient; it's the mathematical tool we use to measure instantaneous change. Whether you are finding the maximum height of a projectile or figuring out how quickly the water level in a tank is rising, differentiation is essential.

In P3, we build on the foundational skills from P1 (polynomials and the basic Chain Rule) and P2 (basic \(e^x\) and \(\ln x\)), introducing powerful new rules to handle much more complex functions—especially those involving products, quotients, implicit relationships, and parametrics.

Key Takeaway from P1/P2 Prerequisites:

You must be comfortable with the standard derivative for powers: \(\frac{d}{dx}(x^n) = nx^{n-1}\).

1. The Core P3 Differentiation Toolkit

In P3, we expand our knowledge of standard derivatives to include the exponential, logarithmic, and a wider range of trigonometric functions.

Standard P3 Derivatives You Must Know

Function, \(f(x)\)	Derivative, \(f'(x)\) or \(\frac{dy}{dx}\)
\(e^x\)	\(e^x\)
\(\ln x\)	\(\frac{1}{x}\)
\(\sin x\)	\(\cos x\)
\(\cos x\)	\(-\sin x\)
\(\tan x\)	\(\sec^2 x\)
\(\tan^{-1} x\)	\(\frac{1}{1 + x^2}\)

(Did you know? The derivative of \(\tan^{-1} x\) is one of the standard results you will be expected to recall and use!)

The Chain Rule (Composites)

The Chain Rule is used when you have a function inside a function. This is often called the Composite Function rule.

If \(y = f(u)\) and \(u = g(x)\), then:

\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

Analogy: The Onion Rule! Start from the outside function, peel one layer off (differentiate it), and then multiply by the derivative of the inner function (the next layer).

Step-by-Step for Composites \(y = e^{f(x)}\):

Example: \(y = e^{3x^2 + 5}\)
Outer function (f(u)): \(e^u\) where \(u = 3x^2 + 5\). Derivative is \(e^u\).
Inner function (u): \(3x^2 + 5\). Derivative is \(\frac{du}{dx} = 6x\).
Result: \(\frac{dy}{dx} = e^{3x^2 + 5} \times (6x) = 6x e^{3x^2 + 5}\).

Common Mistakes to Avoid:

When differentiating \(\ln(f(x))\), the derivative is \(\frac{f'(x)}{f(x)}\). Many students forget to multiply by the derivative of the inner function, \(f'(x)\)!
e.g., If \(y = \ln(2x^3)\), then \(\frac{dy}{dx} = \frac{6x^2}{2x^3} = \frac{3}{x}\).

Key Takeaway (Section 1)

Mastering the Chain Rule is fundamental. Always identify the inner function and multiply by its derivative.

2. Advanced Differentiation Techniques

P3 requires three specialized techniques to differentiate functions involving products, quotients, or non-standard variable relationships.

2.1 The Product Rule

Use this rule when \(y\) is the product of two separate functions of \(x\). If \(y = uv\), where \(u\) and \(v\) are functions of \(x\):

\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)

Mnemonic: "U Dee V Plus V Dee U". Keep one function constant while differentiating the other, then switch roles and add the results.

Step-by-Step Example: \(y = x^2 \sin(3x)\)

Identify \(u\) and \(v\):
- \(u = x^2 \implies \frac{du}{dx} = 2x\)
- \(v = \sin(3x) \implies \frac{dv}{dx} = 3 \cos(3x)\) (using the Chain Rule)
Apply the rule:
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x^2 (3 \cos(3x)) + \sin(3x) (2x)\)
Simplify:
\(\frac{dy}{dx} = 3x^2 \cos(3x) + 2x \sin(3x)\)

2.2 The Quotient Rule

Use this rule when \(y\) is a fraction where both the numerator and denominator are functions of \(x\). If \(y = \frac{u}{v}\):

\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Mnemonic (The Rhyme!): "Low D High, minus High D Low, Square the bottom, and off you go!" (Where Low=\(v\) and High=\(u\)).

Warning! The order matters here because of the minus sign! Start with \(v\) (Low) every time.

2.3 Implicit Differentiation

This technique is necessary when the equation relating \(x\) and \(y\) is complex, and it's difficult or impossible to isolate \(y\) (e.g., \(x^2 + y^2 = xy + 7\)).

When differentiating an implicit function, you treat every term involving \(y\) as a composite function of \(x\).

Rule: When differentiating a term containing \(y\), first differentiate it normally with respect to \(y\), and then immediately multiply by \(\frac{dy}{dx}\) (using the Chain Rule).
Example: Differentiating \(y^3\) with respect to \(x\):
\(\frac{d}{dx} (y^3) = 3y^2 \times \frac{dy}{dx}\)
Example: Differentiating \(xy\) with respect to \(x\): (Use the Product Rule!)
- \(u = x \implies \frac{du}{dx} = 1\)
- \(v = y \implies \frac{dv}{dx} = 1 \times \frac{dy}{dx} = \frac{dy}{dx}\)
- Result: \(\frac{d}{dx} (xy) = x \frac{dy}{dx} + y(1)\)

2.4 Parametric Differentiation

Parametric equations describe a curve using a third variable, often \(t\) (or \(\theta\)). (Imagine \(x\) and \(y\) are both functions of time \(t\). To find the gradient \(\frac{dy}{dx}\), we link their rates of change.)

If \(x = f(t)\) and \(y = g(t)\), the derivative is found by:

\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)

Step-by-Step Process:

Calculate \(\frac{dx}{dt}\) (derivative of \(x\) with respect to \(t\)).
Calculate \(\frac{dy}{dt}\) (derivative of \(y\) with respect to \(t\)).
Divide \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\) to get the final gradient \(\frac{dy}{dx}\) in terms of \(t\).
If the question asks for the gradient at a specific point, substitute the corresponding value of \(t\).

Key Takeaway (Section 2)

The Product and Quotient Rules require careful application of formula memorization. Implicit and Parametric differentiation rely fundamentally on the Chain Rule principle.

3. Applications of Differentiation in P3

3.1 Tangents and Normals

The first derivative \(\frac{dy}{dx}\) always gives the gradient of the tangent to the curve at a specific point.

Gradient of Tangent (\(m_T\)): \(m_T = \frac{dy}{dx}\)
Gradient of Normal (\(m_N\)): The normal is perpendicular to the tangent. \(m_N = -\frac{1}{m_T}\)

Once you have the gradient \(m\) and a point \((x_1, y_1)\), use the straight line formula: \(y - y_1 = m(x - x_1)\).

3.2 Connected Rates of Change

This involves using the Chain Rule to relate two rates of change that are connected through a third variable. This is a common application question in P3.

If you want to find the rate \(\frac{dA}{dt}\) and you know \(\frac{dA}{dr}\) and \(\frac{dr}{dt}\), you can 'link' them:

\(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\)

(Analogy: If you know how fast a sphere's radius (\(r\)) is changing over time (\(\frac{dr}{dt}\)), and you know how the volume (\(V\)) changes with the radius (\(\frac{dV}{dr}\)), you can find how fast the volume is changing over time (\(\frac{dV}{dt}\)).)

Remember to check units! Rates are often given "per second" or "per minute," so time \(t\) must be consistent.

3.3 Stationary Points (Maxima and Minima)

Stationary points occur where the gradient is zero: \(\frac{dy}{dx} = 0\).

To determine the nature of a stationary point (maximum or minimum), you use the Second Derivative Test:

Calculate the second derivative: \(\frac{d^2y}{dx^2}\).
Substitute the \(x\)-coordinate of the stationary point into \(\frac{d^2y}{dx^2}\).

Memory Aid for Second Derivative:

If \(\frac{d^2y}{dx^2} > 0\) (Positive/Happy face), it's a Minimum point.

If \(\frac{d^2y}{dx^2} < 0\) (Negative/Sad face), it's a Maximum point.

3.4 Increasing and Decreasing Functions

The first derivative also tells us about the function's overall behaviour:

Function is Increasing if \(\frac{dy}{dx} > 0\).
Function is Decreasing if \(\frac{dy}{dx} < 0\).

For example, to find the interval where \(y\) is increasing, you solve the inequality \(\frac{dy}{dx} > 0\).

Quick P3 Differentiation Review

This P3 chapter demands precision. Don't worry if the rules seem complicated—practice makes perfect! Here are the core P3 rules one more time:

Product Rule (uv): \(u v' + v u'\)

Quotient Rule (\(\frac{u}{v}\)): \(\frac{v u' - u v'}{v^2}\)

Implicit: Differentiate terms with \(y\) normally, then multiply by \(\frac{dy}{dx}\).

Parametric: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)

Rate of Change (Chain Rule): \(\frac{A}{C} = \frac{A}{B} \times \frac{B}{C}\)

Keep your algebra tidy and memorize those standard derivatives. You've got this!

References

Cambridge International AS & A Level Mathematics (9709) — Cambridge Assessment International Education

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the standard derivatives for \( e^x \), \( \ln x \), and \( \tan^{-1} x \) in P3?

\( \frac{d}{dx}(e^x) = e^x \), \( \frac{d}{dx}(\ln x) = \frac{1}{x} \), and \( \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2} \).

How is the Chain Rule applied to composite functions?

\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \), where \( u \) is the inner function.

What is the Product Rule formula for differentiating \( y = uv \)?

\( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

What is the Quotient Rule formula for differentiating \( y = \frac{u}{v} \)?

\( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

What is the key rule for differentiating terms involving \( y \) during implicit differentiation?

Differentiate the term with respect to \( y \) and then multiply by \( \frac{dy}{dx} \).

How do you find the gradient \( \frac{dy}{dx} \) from parametric equations?

\( \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} \).

How do you calculate the gradient of a normal to a curve?

The gradient of the normal is the negative reciprocal of the tangent gradient: \( m_N = -\frac{1}{dy/dx} \).

What is the formula for relating two rates of change using a third variable?

\( \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} \).

How can you use the second derivative to determine if a stationary point is a maximum or minimum?

If \( \frac{d^2y}{dx^2} > 0 \), it is a minimum; if \( \frac{d^2y}{dx^2} < 0 \), it is a maximum.

How is a function determined to be strictly increasing on an interval?

A function is increasing if its first derivative is positive: \( \frac{dy}{dx} > 0 \).

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