Differentiating Powers and the Chain Rule (9709 Pure Mathematics 1)

Introduction

Differentiation is one of the most heavily examined topics in 9709 Pure Mathematics 1. This note focuses on the two core differentiation skills assessed in objective 1.7: the power rule (extended to all rational powers) and the chain rule for composite functions.

Together, these tools allow you to differentiate a wide class of functions — polynomials, roots, negative powers, and functions-of-a-function — all of which appear routinely in exam questions on gradients, tangents, normals, stationary points, and rates of change.

Core Concept

The Power Rule for $x^n$ (any rational $n$ )

You already know that differentiating $x^n$ gives $nx^{n-1}$ . The crucial extension here is that $n$ can be any rational number — positive, negative, or fractional. This means you can differentiate expressions involving roots (e.g. $\sqrt{x} = x^{1/2}$ ) and reciprocals (e.g. $\tfrac{1}{x^2} = x^{-2}$ ) in exactly the same way as ordinary polynomials.

Constant multiples, sums, and differences follow directly: you differentiate term by term, pulling constants through.

The Chain Rule for Composite Functions

A composite function has the form $y = f(g(x))$ — a function of a function. The chain rule states:

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

where $u = g(x)$ is the inner function. In practice, this means:

Identify the outer function and the inner function.
Differentiate the outer function (treating the inner as the variable).
Multiply by the derivative of the inner function.

A compact direct form: if $y = [f(x)]^n$ , then

\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)

This covers brackets raised to a rational power — the most common chain rule form in P1.

Key Formulae & Definitions

\frac{d}{dx}\left(x^n\right) = nx^{n-1}, \quad n \in \mathbb{Q}

\frac{d}{dx}\left(kx^n\right) = knx^{n-1}, \quad k \text{ constant}

\frac{d}{dx}\left[f(x) \pm g(x)\right] = f'(x) \pm g'(x)

Chain Rule:

\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)

Bracket form:

\frac{d}{dx}\left[(ax+b)^n\right] = n(ax+b)^{n-1} \cdot a

Useful rewrites before differentiating:

Original form	Rewrite as
$\sqrt{x}$	$x^{1/2}$
$\dfrac{1}{x^n}$	$x^{-n}$
$\sqrt[3]{x^2}$	$x^{2/3}$
$\dfrac{1}{\sqrt{x}}$	$x^{-1/2}$

Worked Examples

Example 1 — Power Rule with Sums and Differences

Differentiate $y = 3x^4 - \dfrac{5}{x^2} + 4\sqrt{x} - 7$ .

Step 1 — Rewrite every term using rational indices:

y = 3x^4 - 5x^{-2} + 4x^{1/2} - 7

Step 2 — Apply the power rule term by term:

\frac{dy}{dx} = 3 \cdot 4x^{3} - 5 \cdot (-2)x^{-3} + 4 \cdot \tfrac{1}{2}x^{-1/2} - 0

\frac{dy}{dx} = 12x^{3} + 10x^{-3} + 2x^{-1/2}

Step 3 — Write tidily (leave in index form, or convert back if the question requires):

\frac{dy}{dx} = 12x^3 + \frac{10}{x^3} + \frac{2}{\sqrt{x}}

Justification: The constant $-7$ disappears (derivative of a constant is zero). Each term is treated independently by linearity.

Example 2 — Chain Rule with a Bracket to a Fractional Power

Differentiate $y = \sqrt{3x^2 - 5x + 1}$ with respect to $x$ .

Step 1 — Rewrite using a rational index:

y = (3x^2 - 5x + 1)^{1/2}

Step 2 — Identify outer and inner functions:

Outer: $u^{1/2}$ , where $u = 3x^2 - 5x + 1$
Inner: $u = 3x^2 - 5x + 1$

Step 3 — Differentiate using the chain rule:

\frac{dy}{dx} = \frac{1}{2}(3x^2 - 5x + 1)^{-1/2} \cdot \frac{d}{dx}(3x^2 - 5x + 1)

Step 4 — Compute the derivative of the inner function:

\frac{du}{dx} = 6x - 5

Step 5 — Combine:

\frac{dy}{dx} = \frac{1}{2}(3x^2 - 5x + 1)^{-1/2} \cdot (6x - 5)

\boxed{\frac{dy}{dx} = \frac{6x-5}{2\sqrt{3x^2 - 5x + 1}}}

Justification: The negative index from the outer differentiation is immediately rewritten as a denominator for clarity — examiners reward tidy, unsimplified or simplified answers in equivalent form.

Example 3 — Chain Rule with a Negative Integer Power

Differentiate $y = \dfrac{4}{(2x-3)^3}$ .

Step 1 — Rewrite:

y = 4(2x-3)^{-3}

Step 2 — Apply chain rule:

\frac{dy}{dx} = 4 \cdot (-3)(2x-3)^{-4} \cdot \frac{d}{dx}(2x - 3)

Step 3 — Inner derivative:

\frac{d}{dx}(2x-3) = 2

Step 4 — Combine:

\frac{dy}{dx} = 4 \cdot (-3) \cdot 2 \cdot (2x-3)^{-4} = -24(2x-3)^{-4}

\boxed{\frac{dy}{dx} = -\frac{24}{(2x-3)^4}}

Common Mistakes & Examiner Pitfalls

Forgetting to multiply by the inner derivative. Writing $\dfrac{d}{dx}(3x+1)^5 = 5(3x+1)^4$ scores no marks — the factor of $3$ must appear.
Failing to rewrite before differentiating. Attempting to differentiate $\dfrac{1}{x^3}$ or $\sqrt[4]{x}$ without converting to index form leads to errors. Always rewrite first.
Subtracting 1 from the wrong index. With $x^{-2}$ , the new index is $-3$ , not $-1$ . Keep careful track of negative indices.
Losing the negative sign. When differentiating $x^{-n}$ , the coefficient becomes $-n$ (negative). This is a very common slip.
Treating $\sqrt{f(x)}$ as if it were $\sqrt{x}$ scaled. The chain rule is essential; only $\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}$ is the basic result — a non-trivial inner function always requires the chain rule multiplier.
Leaving answers with negative or fractional indices when the question says "simplify" or when evaluating at a point. Check the demand of the question.

Practice Questions

Q1. Differentiate $y = 5x^3 - \dfrac{3}{\sqrt{x}} + 2x - 9$ .

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Rewrite: $y = 5x^3 - 3x^{-1/2} + 2x - 9$

\frac{dy}{dx} = 15x^2 - 3 \cdot \left(-\tfrac{1}{2}\right)x^{-3/2} + 2

\frac{dy}{dx} = 15x^2 + \frac{3}{2}x^{-3/2} + 2 = 15x^2 + \frac{3}{2x^{3/2}} + 2

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Q2. Find $\dfrac{dy}{dx}$ when $y = (5x^2 - 3)^6$ .

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Let $u = 5x^2 - 3$ , so $y = u^6$ .

\frac{dy}{dx} = 6(5x^2-3)^5 \cdot \frac{d}{dx}(5x^2 - 3)

= 6(5x^2-3)^5 \cdot 10x

\frac{dy}{dx} = 60x(5x^2-3)^5

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Q3. A curve has equation $y = \dfrac{2}{\sqrt{4x-1}}$ . Find the gradient of the curve at the point where $x = \dfrac{5}{4}$ .

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Rewrite: $y = 2(4x-1)^{-1/2}$

Apply chain rule:

\frac{dy}{dx} = 2 \cdot \left(-\tfrac{1}{2}\right)(4x-1)^{-3/2} \cdot 4 = -4(4x-1)^{-3/2}

At $x = \tfrac{5}{4}$ : $4x - 1 = 5 - 1 = 4$

\frac{dy}{dx} = -4 \cdot 4^{-3/2} = -4 \cdot \frac{1}{8} = -\frac{1}{2}

The gradient at that point is $-\dfrac{1}{2}$ .

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Q4. Differentiate $y = x^2 + \dfrac{1}{(3-2x)^4}$ and find the $x$ -value(s) where $\dfrac{dy}{dx} = 0$ .

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Rewrite: $y = x^2 + (3-2x)^{-4}$

\frac{dy}{dx} = 2x + (-4)(3-2x)^{-5} \cdot (-2) = 2x + 8(3-2x)^{-5}

Set equal to zero:

2x + \frac{8}{(3-2x)^5} = 0

2x(3-2x)^5 = -8

x(3-2x)^5 = -4

This equation has no simple closed-form solution by elementary algebra; at P1 level, questions are typically set so that a specific value is sought numerically or an approximate value is acceptable. Check the question context for whether exact or decimal form is required.

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Q5. Given $f(x) = \left(x^3 + \dfrac{1}{x}\right)^{1/2}$ , find $f'(x)$ , leaving your answer in simplified surd form.

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Let $u = x^3 + x^{-1}$ , so $f(x) = u^{1/2}$ .

f'(x) = \frac{1}{2}\left(x^3 + x^{-1}\right)^{-1/2} \cdot \frac{d}{dx}\left(x^3 + x^{-1}\right)

\frac{du}{dx} = 3x^2 - x^{-2}

f'(x) = \frac{3x^2 - x^{-2}}{2\sqrt{x^3 + x^{-1}}}

Multiplying numerator and denominator by $x^2$ to clear the inner negative index:

f'(x) = \frac{3x^4 - 1}{2x^2\sqrt{x^3 + x^{-1}}}

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Connections

Prerequisite — The Derivative and Its Notation: This note builds directly on the definition of $\dfrac{dy}{dx}$ and the basic idea that the derivative gives the gradient of the tangent. Fluency with index notation is assumed throughout.

Next: Tangents and Normals — You will apply these differentiation skills to evaluate $\dfrac{dy}{dx}$ at a specific point and hence find the equations of tangent and normal lines.

Next: Stationary Points and Curve Sketching — Setting $\dfrac{dy}{dx} = 0$ and classifying results requires exactly the differentiation techniques in this note.

Next: Differentiation of Products and Quotients (A2 Further Mathematics / extension) — The chain rule is a building block for the product and quotient rules encountered in more advanced work.

Connected: Integration of $x^n$ — The reverse process (the power rule for integration) mirrors the differentiation power rule exactly; recognising this symmetry aids in checking answers.

Introduction

Core Concept

The Power Rule for $x^n$ (any rational $n$ )

The Chain Rule for Composite Functions

Key Formulae & Definitions

Worked Examples

Example 1 — Power Rule with Sums and Differences

Example 2 — Chain Rule with a Bracket to a Fractional Power

Example 3 — Chain Rule with a Negative Integer Power

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Introduction

Core Concept

The Power Rule for xnx^nxn (any rational nnn)

The Chain Rule for Composite Functions

Key Formulae & Definitions

Worked Examples

Example 1 — Power Rule with Sums and Differences

Example 2 — Chain Rule with a Bracket to a Fractional Power

Example 3 — Chain Rule with a Negative Integer Power

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

The Power Rule for $x^n$ (any rational $n$ )