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Module 3 · L04 of 15 ~40 min +95 XP available

Differentiation Rules

Differentiating from first principles is powerful, but tedious. What if there were shortcuts? In this lesson, you will learn the three fundamental rules, power, constant multiple, and sum/difference, that let you differentiate polynomials in seconds instead of minutes.

Today's hook, Consider $f(x) = x^3 + 2x$. If you had to find its derivative using first principles, you would need to expand $(x+h)^3$, simplify, cancel $h$, and take a limit. That takes several minutes. Do you think there might be a faster way to differentiate terms like $x^3$ and $2x$ individually?

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Consider $f(x) = x^3 + 2x$. If you had to find its derivative using first principles, you would need to expand $(x+h)^3$, simplify, cancel $h$, and take a limit. That takes several minutes. Do you think there might be a faster way to differentiate terms like $x^3$ and $2x$ individually? What pattern do you notice from derivatives you have already calculated?

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The three differentiation rules

Work through the core explanation before applying it.

The three differentiation rules

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There are only three fundamental rules you need to differentiate any polynomial in seconds. Each rule is a shortcut that avoids first principles. Together they let you differentiate term-by-term, no matter how many terms the function has.

The power rule says: bring the exponent down as a coefficient, then subtract one from the exponent. The constant multiple rule says: constants slide past the derivative unchanged. The sum/difference rule says: differentiate each term separately and add or subtract the results.

Differentiation rules summary as a comparison table of rule name to formula

Power · Constant · Sum

Power Rule

$$\frac{d}{dx}\bigl(x^n\bigr) = nx^{n-1}$$ for all real $n$. Bring the power down, subtract one from the exponent.

Constant Multiple

$$\frac{d}{dx}\bigl(cf(x)\bigr) = c \cdot f'(x)$$ Constants slide straight past the derivative unchanged.

Sum / Difference

$$\frac{d}{dx}\bigl(f(x) \pm g(x)\bigr) = f'(x) \pm g'(x)$$ Differentiate term by term, then combine.

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What you will master

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

What you will master

Know

Key facts

The power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
The constant multiple and sum/difference rules
How to rewrite roots and reciprocals as powers

Understand

Concepts

Why the power rule is consistent with first principles
How differentiation is a linear operation
When to rewrite expressions before differentiating

Can do

Skills

Differentiate polynomials efficiently using the rules
Rewrite radicals and fractions as powers to differentiate
Find derivatives at specific points

Key terms

Power RuleIf $f(x) = x^n$, then $f'(x) = nx^{n-1}$ for all real $n$.

Constant Multiple Rule$\frac{d}{dx}(cf(x)) = c \cdot f'(x)$, constants pass through unchanged.

Sum Rule$\frac{d}{dx}(f(x) + g(x)) = f'(x) + g'(x)$, differentiate term by term.

Difference Rule$\frac{d}{dx}(f(x) - g(x)) = f'(x) - g'(x)$, same as sum but with subtraction.

CoefficientThe numerical factor multiplying a variable term, e.g. the 3 in $3x^2$.

PolynomialA sum of terms of the form $ax^n$ where $n$ is a non-negative integer.

DifferentiationThe process of finding the derivative of a function.

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How differentiation rules work

Work through the core explanation before applying it.

How differentiation rules work

core concept

Differentiation rules are shortcuts that avoid the lengthy first-principles definition. The power rule is the engine: if you can write every term as $x^n$, you can differentiate it instantly.

The power rule can be justified from first principles for positive integers using the binomial expansion of $(x+h)^n$. When you expand, subtract $x^n$, divide by $h$, and take the limit as $h \to 0$, every term except $nx^{n-1}$ vanishes. For negative and fractional exponents, the rule still holds, it is valid for all real numbers $n$.

The constant multiple and sum rules follow from the linearity of limits. Since the derivative is defined as a limit, and limits distribute over addition and scalar multiplication, differentiation does too. This means you can differentiate term by term, pulling constants out as you go.

Many functions need rewriting before the power rule applies:

$$\sqrt{x} = x^{1/2}, \quad \frac{1}{x^2} = x^{-2}, \quad \frac{3}{\sqrt{x}} = 3x^{-1/2}$$

Always rewrite in index form first, then apply the power rule to each term.

Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$, bring the power down, subtract 1 from the exponent; Constant multiple: $\frac{d}{dx}(cf) = cf'$, constants slide through

Pause, copy the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ (bring down, reduce by 1), the constant multiple rule, and the "rewrite in index form first" reminder into your book.

Quick check: Using the power rule, $\frac{d}{dx}(x^5) = $?

Worked examples · 3 in a row, reveal as you go

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Follow the reasoning through complete worked solutions.

PROBLEM 1 · DIFFERENTIATE A POLYNOMIAL

Differentiate $f(x) = 3x^4 - 2x^3 + 5x - 7$.

Apply the power rule to each term: $\frac{d}{dx}(3x^4) = 3 \cdot 4x^3 = 12x^3$

Multiply the coefficient by the exponent, then subtract 1 from the exponent.

PROBLEM 2 · REWRITE AS POWERS FIRST

Differentiate $f(x) = \sqrt{x} + \frac{1}{x^2}$.

Rewrite using index notation: $f(x) = x^{1/2} + x^{-2}$

Always convert radicals and reciprocals to powers before differentiating.

PROBLEM 3 · FIND A DERIVATIVE AT A POINT

Find $f'(2)$ for $f(x) = x^3 - 4x + 1$.

$f'(x) = 3x^2 - 4$

Differentiate the function using the power rule and constant multiple rule.

Quick check: True or false, the derivative of the constant 7 is 7.

Common errors · the 3 traps that cost marks

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

Trap 01

Forgetting to subtract 1 from the exponent

When applying the power rule, students sometimes multiply by the power but forget to reduce it by one. For $x^5$, the derivative is $5x^4$, not $5x^5$. Both operations, bringing the power down AND subtracting one, are required.

Trap 02

Treating coefficients incorrectly

Some students multiply the coefficient by the power and then multiply again, getting $3x^4 \to 12x^4$ instead of $12x^3$. The coefficient stays as a coefficient after the power rule is applied: $\frac{d}{dx}(3x^4) = 3 \cdot 4x^3 = 12x^3$.

Trap 03

Not rewriting radicals and reciprocals as powers first

Trying to differentiate $\sqrt{x}$ or $\frac{1}{x^2}$ directly without rewriting as $x^{1/2}$ or $x^{-2}$ leads to errors. Always rewrite in index form before applying the power rule: $\sqrt{x} = x^{1/2}$, $\frac{1}{x^3} = x^{-3}$, $\frac{2}{\sqrt{x}} = 2x^{-1/2}$.

Odd one out: Three of these can be differentiated immediately using the power rule. Which one must be rewritten first?

Quick-fire practice · 5 problems

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Run the quick drill and copy the summary into your book.

Differentiate $f(x) = x^5$.

Show answer

$f'(x) = 5x^4$
Using the power rule: bring the 5 down and subtract 1 from the exponent.

Differentiate $f(x) = 4x^3 - 3x^2 + 2x - 1$.

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$f'(x) = 12x^2 - 6x + 2$
Term by term: $4 \cdot 3x^2 = 12x^2$, $-3 \cdot 2x = -6x$, $2 \cdot 1 = 2$, $-1 \to 0$.

Differentiate $f(x) = \frac{1}{x^3}$.

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$f'(x) = -\frac{3}{x^4}$
Rewrite as $x^{-3}$, then $\frac{d}{dx}(x^{-3}) = -3x^{-4} = -\frac{3}{x^4}$.

Differentiate $f(x) = 3\sqrt{x} - \frac{2}{x}$.

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$f'(x) = \frac{3}{2\sqrt{x}} + \frac{2}{x^2}$
Rewrite as $3x^{1/2} - 2x^{-1}$. Then $3 \cdot \frac{1}{2}x^{-1/2} - 2(-1)x^{-2} = \frac{3}{2\sqrt{x}} + \frac{2}{x^2}$.

Find $f'(1)$ for $f(x) = 2x^4 - 3x + 5$.

Show answer

$f'(1) = 5$
$f'(x) = 8x^3 - 3$, so $f'(1) = 8 - 3 = 5$.

Fill the blanks: drag each token into the matching blank.

nx^(n-1) index form zero term by term

The power rule states $\frac{d}{dx}(x^n) = $ ___. Before applying the power rule to radicals, always rewrite in ___. The derivative of any constant is ___. The sum rule lets you differentiate ___.

Two truths and a lie: Identify the false statement about differentiation rules.

Revisit your thinking

Return to your original answer from Section 01. You were asked about $f(x) = x^3 + 2x$.

Using the power rule, we differentiate each term separately. The derivative of $x^3$ is $3x^2$, and the derivative of $2x$ is $2$. So:

$$f'(x) = 3x^2 + 2$$

This takes seconds rather than minutes. This is why the differentiation rules are so powerful: they preserve the rigour of calculus while making computation practical.

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Multiple choice

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Multiple choice

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Short answer

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Short answer

Apply Band 4

Differentiate a mixed function

Differentiate $f(x) = x^4 - 3x^2 + \frac{2}{x} - \sqrt{x}$. Show all working. 4 marks

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Step 1: Rewrite in index form.

$$f(x) = x^4 - 3x^2 + 2x^{-1} - x^{1/2}$$

Step 2: Differentiate term by term.

$$\frac{d}{dx}(x^4) = 4x^3, \quad \frac{d}{dx}(-3x^2) = -6x$$

$$\frac{d}{dx}(2x^{-1}) = -2x^{-2} = -\frac{2}{x^2}, \quad \frac{d}{dx}(-x^{1/2}) = -\frac{1}{2}x^{-1/2} = -\frac{1}{2\sqrt{x}}$$

Answer: $f'(x) = \mathbf{4x^3 - 6x - \frac{2}{x^2} - \frac{1}{2\sqrt{x}}}$.

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Gradient of a tangent

Find the gradient of the tangent to $y = x^3 - 2x^2 + 4$ at $x = -1$. 3 marks

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Step 1: Differentiate.

$$\frac{dy}{dx} = 3x^2 - 4x$$

Step 2: Substitute $x = -1$.

$$\frac{dy}{dx}\bigg|_{x=-1} = 3(-1)^2 - 4(-1) = 3 + 4 = 7$$

Answer: The gradient is $\mathbf{7}$.

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Prove the power rule

Prove the power rule for positive integer exponents using the binomial expansion in the first principles definition. 4 marks

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First principles definition:

$$f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$$

Expand $(x+h)^n$ using the binomial theorem:

$$(x+h)^n = x^n + nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n$$

Subtract $x^n$ and divide by $h$:

$$\frac{(x+h)^n - x^n}{h} = nx^{n-1} + \binom{n}{2}x^{n-2}h + \cdots + h^{n-1}$$

Take the limit as $h \to 0$:

$$f'(x) = nx^{n-1}$$

Hence proved: $\frac{d}{dx}(x^n) = \mathbf{nx^{n-1}}$ for positive integers $n$.

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Module overview · Maths Advanced · Checkpoint 1