Simplifying and Differentiating Revision Notes | GCSE AQA Higher Further Maths

1. Why Differentiation Needs a Clean Expression First

You already know how to find the gradient of a straight line. Differentiation does the same job for curves — it tells you the gradient at any point. The catch is that the rule only works cleanly when your expression is written as a sum of terms in the form $kx^n$.

Many exam questions deliberately give you an expression with brackets or fractions. Your first job is always to rewrite it into separate $kx^n$ terms. Once that is done, differentiating becomes mechanical and reliable.

Before you can use this rule, the expression must be in the right shape. That means no brackets left unexpanded, no $x$ terms sitting in a denominator, and no roots left as $\sqrt{x}$ — rewrite them using index notation first.

2. Simplifying Before You Differentiate

There are three common situations where you must simplify before differentiating. Each one has a specific rewriting move.

• Brackets: expand fully so every term is separate, e.g. $x(x^2 + 3) = x^3 + 3x$
• Fractions with $x$ in the denominator: use negative indices, e.g. $\frac{5}{x^2} = 5x^{-2}$
• Roots: use fractional indices, e.g. $\sqrt{x} = x^{1/2}$, $\sqrt[3]{x} = x^{1/3}$

Worked example: Expanding brackets before differentiating

Find $\frac{dy}{dx}$ when $y = x(3x^2 - 4x + 1)$.

1. $y = x \cdot 3x^2 - x \cdot 4x + x \cdot 1 = 3x^3 - 4x^2 + x$ — Multiply each term inside the bracket by $x$. Now every term is in $kx^n$ form. [M1]
2. $\frac{d}{dx}(3x^3) = 3 \times 3 \cdot x^{3-1} = 9x^2$ — Multiply coefficient 3 by power 3, then reduce the power by 1.
3. $\frac{d}{dx}(-4x^2) = -4 \times 2 \cdot x^{2-1} = -8x$ — Multiply $-4$ by $2$, reduce power from 2 to 1.
4. $\frac{d}{dx}(x) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$ — $x = 1 \cdot x^1$, so the derivative is $1 \times 1 \cdot x^0 = 1$. [A1]
5. $\frac{dy}{dx} = 9x^2 - 8x + 1$ — Collect all differentiated terms. This is the gradient function for the curve.

Answer: $\frac{dy}{dx} = 9x^2 - 8x + 1$

3. Fractions and Negative Indices

When $x$ appears in the denominator of a fraction, the power rule cannot be applied directly. Rewrite using a negative index first: $\frac{k}{x^n} = kx^{-n}$. After that, the rule works exactly as before.

Worked example: Differentiating an expression with a fraction

Find $\frac{dy}{dx}$ when $y = \frac{6x^3 - 4}{x^2}$.

1. $y = \frac{6x^3}{x^2} - \frac{4}{x^2} = 6x^{3-2} - 4x^{-2} = 6x - 4x^{-2}$ — Divide each term in the numerator by $x^2$ separately. Use index laws: $x^3 \div x^2 = x^1$. [M1]
2. $\frac{d}{dx}(6x) = 6$ — $6x = 6x^1$, so the derivative is $6 \times 1 \cdot x^0 = 6$.
3. $\frac{d}{dx}(-4x^{-2}) = -4 \times (-2) \cdot x^{-2-1} = 8x^{-3}$ — Multiply $-4$ by $-2$ to get $+8$, then reduce the power from $-2$ to $-3$. [M1]
4. $\frac{dy}{dx} = 6 + 8x^{-3}$ — You may also write this as $6 + \frac{8}{x^3}$. Both forms are acceptable. [A1]

Answer: $\frac{dy}{dx} = 6 + 8x^{-3}$ (or equivalently $6 + \frac{8}{x^3}$)

The graph above shows $y = 6x - 4x^{-2}$ (the original function) and $\frac{dy}{dx} = 6 + 8x^{-3}$ (the derivative). Notice that where the original curve is steepest, the derivative takes its largest values — this is exactly what differentiation measures.

4. Putting It Together — Mixed Expressions

Exam questions often combine both challenges at once: brackets and fractions in the same expression. The strategy is always the same — fully simplify into $kx^n$ terms, then differentiate each term using the power rule.

Worked example: Simplifying a product divided by $x$ before differentiating

Find $\frac{dy}{dx}$ when $y = \frac{(x+3)(x-1)}{x}$.

1. $(x+3)(x-1) = x^2 - x + 3x - 3 = x^2 + 2x - 3$ — Use FOIL or the grid method to expand the brackets fully. [M1]
2. $y = \frac{x^2 + 2x - 3}{x} = x + 2 - 3x^{-1}$ — $\frac{x^2}{x} = x$, $\frac{2x}{x} = 2$, $\frac{-3}{x} = -3x^{-1}$. Now every term is in $kx^n$ form. [M1]
3. $\frac{d}{dx}(x) = 1$ — $x = x^1$, so derivative is $1 \times x^0 = 1$.
4. $\frac{d}{dx}(2) = 0$ — Constants always differentiate to zero — there is no $x$ to apply the rule to.
5. $\frac{d}{dx}(-3x^{-1}) = -3 \times (-1) \cdot x^{-1-1} = 3x^{-2}$ — Multiply $-3$ by $-1$ to get $+3$, reduce power from $-1$ to $-2$. [M1]
6. $\frac{dy}{dx} = 1 + 0 + 3x^{-2} = 1 + 3x^{-2}$ — Or equivalently $1 + \frac{3}{x^2}$. [A1]

Answer: $\frac{dy}{dx} = 1 + 3x^{-2}$ (or $1 + \frac{3}{x^2}$)

5. Practice — Apply the Full Method

These problems cover the full range of simplification types. For each one, write out the simplified form before differentiating — do not skip that step, even under time pressure.

Notice that in the third problem the numerator expansion is the critical step — an error there carries through to the wrong derivative. Always check your expansion before moving on.

6. Exam Technique and Common Mistakes

Marks in differentiation questions are split between the method (simplifying correctly) and the answer (applying the power rule correctly). Losing marks on the simplification step is avoidable with a clear written process.

Working through that ordering confirms the key principle: simplify fully into $kx^n$ terms, then apply the power rule term by term. Every differentiation question that involves brackets or fractions follows this same two-stage process.