Basic Differentiation

This lesson covers the power rule, the sum/difference rule, and applications to tangents and normals, as required by the Edexcel A-Level Mathematics specification (9MA0). These are the foundational differentiation skills you will use throughout the course.

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The Power Rule

The most important differentiation rule:

If y = xⁿ, then dy/dx = nxⁿ⁻¹

This works for any real value of n — positive, negative, fractional, or zero.

y	dy/dx
x²	2x
x³	3x²
x⁵	5x⁴
x	1 (since x = x¹)
x⁰ = 1	0 (derivative of a constant)
x⁻¹ = 1/x	-x⁻² = -1/x²
x⁻² = 1/x²	-2x⁻³ = -2/x³
x^(1/2) = √x	(1/2)x^(-1/2) = 1/(2√x)
x^(3/2)	(3/2)x^(1/2) = (3/2)√x

Constant Multiple Rule

If y = kxⁿ (where k is a constant), then:

dy/dx = knxⁿ⁻¹

"The constant stays, differentiate the power as normal."

Worked Example 1

Differentiate: (a) y = 4x³ (b) y = 7x (c) y = -2x⁵ (d) y = (1/3)x⁶

Solution: (a) dy/dx = 4 × 3x² = 12x² (b) dy/dx = 7 × 1 = 7 (c) dy/dx = -2 × 5x⁴ = -10x⁴ (d) dy/dx = (1/3) × 6x⁵ = 2x⁵

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The Sum and Difference Rules

If y = f(x) ± g(x), then:

dy/dx = f'(x) ± g'(x)

"Differentiate each term separately."

Worked Example 2

Differentiate y = 3x⁴ - 2x³ + 5x² - 7x + 1.

Solution: dy/dx = 12x³ - 6x² + 10x - 7

Each term is differentiated independently. The constant (+1) differentiates to 0.

Worked Example 3

Differentiate y = 2x³ + 4/x - 3√x.

Solution: First rewrite in power form: y = 2x³ + 4x⁻¹ - 3x^(1/2)

dy/dx = 6x² + 4(-1)x⁻² - 3(1/2)x^(-1/2) = 6x² - 4x⁻² - (3/2)x^(-1/2) = 6x² - 4/x² - 3/(2√x)

Exam Tip: Before differentiating, always rewrite roots and fractions as powers of x. For example, √x = x^(1/2), 1/x = x⁻¹, 1/x² = x⁻², ∛x = x^(1/3).

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Finding the Gradient at a Point

To find the gradient of a curve at a specific point:

1. Differentiate to find dy/dx.
2. Substitute the x-value of the point.

Worked Example 4

Find the gradient of the curve y = x³ - 4x + 2 at the point (2, 2).

Solution: dy/dx = 3x² - 4

At x = 2: dy/dx = 3(4) - 4 = 12 - 4 = 8

The gradient at (2, 2) is 8.

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Tangents

The tangent to a curve at a point touches the curve and has the same gradient as the curve at that point.

Method for Finding the Tangent

1. Find dy/dx.
2. Substitute the x-value to get the gradient m.
3. Find the y-value if not given (substitute x into the original equation).
4. Use y - y₁ = m(x - x₁) to write the equation.

Worked Example 5

Find the equation of the tangent to y = x² - 3x + 5 at the point where x = 2.

Solution: dy/dx = 2x - 3. At x = 2: m = 2(2) - 3 = 1

y-value: y = 4 - 6 + 5 = 3. Point is (2, 3).

Tangent: y - 3 = 1(x - 2) y = x + 1

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Normals

The normal to a curve at a point is perpendicular to the tangent at that point.

If the tangent has gradient m, the normal has gradient -1/m (the negative reciprocal).

Method for Finding the Normal

1. Find the gradient of the tangent, m.
2. Calculate the normal gradient: -1/m.
3. Use y - y₁ = (-1/m)(x - x₁).

Worked Example 6

Find the equation of the normal to y = x³ at the point (1, 1).

Solution: dy/dx = 3x². At x = 1: m = 3(1)² = 3

Normal gradient: -1/3

Normal: y - 1 = (-1/3)(x - 1) 3(y - 1) = -(x - 1) 3y - 3 = -x + 1 x + 3y = 4 or equivalently y = (-1/3)x + 4/3

Exam Tip: If the tangent gradient is 0 (horizontal tangent), the normal is vertical (equation x = a). If the tangent is vertical (gradient undefined), the normal is horizontal (equation y = b).

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Differentiating Expressions That Need Simplifying First

Before differentiating, you sometimes need to expand brackets or simplify fractions.

Worked Example 7

Differentiate y = (2x + 1)(3x - 4).

Solution: Expand first: y = 6x² - 8x + 3x - 4 = 6x² - 5x - 4

dy/dx = 12x - 5

Worked Example 8

Differentiate y = (x³ + 4x) / x.

Solution: Divide each term by x: y = x² + 4

dy/dx = 2x

Worked Example 9

Differentiate y = (3x² - 1)².

Solution: Expand: y = 9x⁴ - 6x² + 1

dy/dx = 36x³ - 12x = 12x(3x² - 1)

Exam Tip: At this stage, you can only differentiate sums of terms of the form axⁿ. You cannot differentiate products, quotients, or compositions directly — those require the product rule, quotient rule, or chain rule (covered later).

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Increasing and Decreasing Functions

A function is increasing on an interval if dy/dx > 0 throughout that interval. A function is decreasing on an interval if dy/dx < 0 throughout that interval.

Worked Example 10

Show that y = x³ + 3x + 2 is an increasing function for all x.

Solution: dy/dx = 3x² + 3 = 3(x² + 1)

Since x² ≥ 0 for all x, we have x² + 1 ≥ 1 > 0 for all x. Therefore dy/dx = 3(x² + 1) > 0 for all x.

Hence y is an increasing function for all values of x. ∎

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Common Mistakes

Mistake	Correction
d/dx(x) = 0	d/dx(x) = 1, not 0. The derivative of a constant is 0, but x is not a constant
d/dx(5) = 5	d/dx(5) = 0. The derivative of any constant is zero
Differentiating 1/x as -1/x	Rewrite as x⁻¹ first: d/dx(x⁻¹) = -x⁻² = -1/x²
Forgetting to rewrite before differentiating	Always convert √x, 1/x, etc. to power form first
Using product rule when expanding will do	At this stage, expand brackets before differentiating

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Summary

• The power rule: d/dx(xⁿ) = nxⁿ⁻¹ for any real n.
• Constant multiple: d/dx(kxⁿ) = knxⁿ⁻¹.
• Sum/difference: differentiate each term separately.
• To find a tangent, use the gradient at the point with y - y₁ = m(x - x₁).
• The normal is perpendicular to the tangent with gradient -1/m.
• Always rewrite roots, fractions, and products as sums of xⁿ terms before differentiating (at this stage).

A-Level Deep Dive: Basic Differentiation

Spec mapping

Edexcel 9MA0-01 specification section 9 — Differentiation, sub-strands 9.1 and 9.2 covers the derivative of $f(x)$f(x) as the gradient of the tangent to the graph of $y = f(x)$y=f(x) at a general point; differentiate $x^n$xn, for rational values of $n$n, and related constant multiples, sums and differences; differentiate $e^{kx}$ekx and $a^{kx}$akx, $\sin kx$sinkx, $\cos kx$coskx and related sums, differences and constant multiples (refer to the official specification document for exact wording). Section 9 is examined on Paper 1 (Pure Mathematics) and synoptically on Paper 2. The Edexcel formula booklet does not list the standard derivatives $\frac{d}{dx}e^x = e^x$dxd​ex=ex, $\frac{d}{dx}\ln x = 1/x$dxd​lnx=1/x, $\frac{d}{dx}\sin x = \cos x$dxd​sinx=cosx or $\frac{d}{dx}\cos x = -\sin x$dxd​cosx=−sinx — these must be memorised. Synoptic threads include tangent and normal lines (section 7, Coordinate geometry), stationary points (section 9.3), rates of change in modelling (section 9.4), integration as the inverse of differentiation (section 8), and mechanics ($v = \frac{ds}{dt}$v=dtds​, $a = \frac{dv}{dt}$a=dtdv​ in 9MA0-03).

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Worked example with full mark scheme

Question (8 marks): Given that $f(x) = 3x^4 - 5e^x + 4\sin x - \dfrac{2}{x} + \sqrt{x}$f(x)=3x4−5ex+4sinx−x2​+x​ for $x > 0$x>0, find $f'(x)$f′(x), giving each term in its simplest form.

Solution with mark scheme:

Step 1 — rewrite all terms as constant multiples of $x^n$xn or recognised standard forms.

$f(x) = 3x^4 - 5e^x + 4\sin x - 2x^{-1} + x^{1/2}$f(x)=3x4−5ex+4sinx−2x−1+x1/2

M1 — converting $\dfrac{2}{x}$x2​ to $2x^{-1}$2x−1 and $\sqrt{x}$x​ to $x^{1/2}$x1/2. This rewriting step is the gatekeeper for the rest of the question; without it the power rule cannot be applied. Common error: leaving $\dfrac{2}{x}$x2​ untouched and writing the derivative as $\dfrac{2}{1}$12​ or $-\dfrac{2}{x^2}$−x22​ without showing the intermediate $2x^{-1}$2x−1 form.

Step 2 — differentiate the polynomial term $3x^4$3x4.

$\dfrac{d}{dx}(3x^4) = 3 \cdot 4 x^{4-1} = 12x^3$dxd​(3x4)=3⋅4x4−1=12x3

M1 — application of the power rule $\dfrac{d}{dx}(ax^n) = anx^{n-1}$dxd​(axn)=anxn−1 to a positive integer power.

Step 3 — differentiate the exponential term.

$\dfrac{d}{dx}(-5e^x) = -5e^x$dxd​(−5ex)=−5ex

M1 — recognition of $\dfrac{d}{dx}e^x = e^x$dxd​ex=ex and the scalar-multiple rule. The exponential function is the unique non-zero function (up to a multiplicative constant) equal to its own derivative.

Step 4 — differentiate the trigonometric term.

$\dfrac{d}{dx}(4\sin x) = 4\cos x$dxd​(4sinx)=4cosx

A1 — standard derivative $\dfrac{d}{dx}\sin x = \cos x$dxd​sinx=cosx applied with the scalar multiple. Watch the sign: $\dfrac{d}{dx}\cos x = -\sin x$dxd​cosx=−sinx, not $\sin x$sinx.

Step 5 — differentiate the negative-power term.

$\dfrac{d}{dx}(-2x^{-1}) = -2 \cdot (-1) x^{-1-1} = 2x^{-2} = \dfrac{2}{x^2}$dxd​(−2x−1)=−2⋅(−1)x−1−1=2x−2=x22​

M1 — power rule applied with negative $n$n. The double-negative collapsing to a positive coefficient is a routine sign-management point; an A* candidate writes the intermediate $-2 \cdot (-1)$−2⋅(−1) explicitly to make the cancellation auditable.

Step 6 — differentiate the fractional-power term.

$\dfrac{d}{dx}(x^{1/2}) = \dfrac{1}{2} x^{1/2 - 1} = \dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}$dxd​(x1/2)=21​x1/2−1=21​x−1/2=2x​1​

M1 — power rule with rational $n$n. The simplification $x^{-1/2} = \dfrac{1}{\sqrt{x}}$x−1/2=x​1​ converts back to surd form for the final answer.

Step 7 — sum and present.

$f'(x) = 12x^3 - 5e^x + 4\cos x + \dfrac{2}{x^2} + \dfrac{1}{2\sqrt{x}}$f′(x)=12x3−5ex+4cosx+x22​+2x​1​

A1 — fully simplified derivative with each term in its tidiest form (positive indices and surds where appropriate).

Total: 8 marks (M6 A2). The mark scheme rewards each rewritten term separately because Edexcel treats this as testing fluency across all five standard derivative types simultaneously.

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A curve has equation $y = 2x^3 + \dfrac{6}{\sqrt{x}} - \sin x$y=2x3+x​6​−sinx for $x > 0$x>0.

(a) Find $\dfrac{dy}{dx}$dxdy​. (4)

(b) Hence find the gradient of the curve at the point where $x = \dfrac{\pi}{2}$x=2π​, leaving your answer in exact form. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — rewriting $\dfrac{6}{\sqrt{x}}$x​6​ as $6x^{-1/2}$6x−1/2 before differentiating.
• M1 (AO1.1b) — applying the power rule term-by-term: $\dfrac{d}{dx}(2x^3) = 6x^2$dxd​(2x3)=6x2 and $\dfrac{d}{dx}(6x^{-1/2}) = -3x^{-3/2}$dxd​(6x−1/2)=−3x−3/2.
• M1 (AO1.1b) — applying the standard derivative $\dfrac{d}{dx}(-\sin x) = -\cos x$dxd​(−sinx)=−cosx.
• A1 (AO1.1b) — combining: $\dfrac{dy}{dx} = 6x^2 - 3x^{-3/2} - \cos x$dxdy​=6x2−3x−3/2−cosx or equivalent surd form $6x^2 - \dfrac{3}{x\sqrt{x}} - \cos x$6x2−xx​3​−cosx.

(b)

• M1 (AO2.1) — substituting $x = \pi/2$x=π/2 into the derivative and recognising $\cos(\pi/2) = 0$cos(π/2)=0.
• A1 (AO2.5) — exact answer $\dfrac{3\pi^2}{2} - \dfrac{3}{(\pi/2)\sqrt{\pi/2}}$23π2​−(π/2)π/2​3​, kept as the unsimplified exact substitution.

Total: 6 marks split AO1 = 4, AO2 = 2. The "exact form" instruction in (b) means decimals lose the final A1 — a recurring trap on Paper 1.

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Synoptic links

Connects to:

1. Section 9.5 — Chain rule: the next lesson generalises the power rule to $\dfrac{d}{dx}f(g(x)) = f'(g(x)) g'(x)$dxd​f(g(x))=f′(g(x))g′(x). Mastery of basic differentiation is the necessary scaffold; without confidence on $\dfrac{d}{dx}x^n$dxd​xn in isolation, applying it inside a chain is impossible.

2. Section 7 — Coordinate geometry (tangent and normal lines): a tangent at point $(a, f(a))$(a,f(a)) has gradient $f'(a)$f′(a) and equation $y - f(a) = f'(a)(x - a)$y−f(a)=f′(a)(x−a). The normal has gradient $-1/f'(a)$−1/f′(a). Every tangent/normal question on Paper 1 begins with a basic differentiation.

3. Section 9.3 — Stationary points: setting $f'(x) = 0$f′(x)=0 to find critical points and using the second derivative $f''(x)$f′′(x) to classify them as maxima, minima or inflexions. The classification criterion $f''(x) > 0 \Rightarrow$f′′(x)>0⇒ minimum, $f''(x) < 0 \Rightarrow$f′′(x)<0⇒ maximum is an extension of basic differentiation applied twice.

4. Section 8 — Integration: integration is anti-differentiation. $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$∫xndx=n+1xn+1​+C for $n \neq -1$n=−1 is the power rule run backwards. The exception $n = -1$n=−1 leads to $\int \dfrac{1}{x}\,dx = \ln|x| + C$∫x1​dx=ln∣x∣+C, which mirrors $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$dxd​lnx=x1​.

5. Modelling rates of change (section 9.4 and Mechanics 9MA0-03): $\dfrac{dy}{dt}$dtdy​ is the instantaneous rate of change of $y$y with respect to $t$t. Velocity is $v = \dfrac{ds}{dt}$v=dtds​, acceleration is $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$a=dtdv​=dt2d2s​. Every kinematics question with non-uniform acceleration is a basic-differentiation problem in disguise.

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Mark-scheme literacy

Basic differentiation questions on 9MA0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	70–85%	Rewriting roots and reciprocals as fractional/negative indices, applying the power rule term-by-term, recalling standard derivatives $e^x$ex, $\sin x$sinx, $\cos x$cosx, $\ln x$lnx
AO2 (reasoning / interpretation)	10–25%	Choosing the most efficient simplification, presenting derivatives in surd or fractional form as the question demands, justifying domain restrictions (e.g. $x > 0$x>0 for $\ln x$lnx)
AO3 (problem-solving)	0–10%	Modelling problems where the candidate must construct the function before differentiating it

Examiner-rewarded phrasing: "rewriting as $2x^{-1}$2x−1 before applying the power rule"; "by the standard derivative $\dfrac{d}{dx}\sin x = \cos x$dxd​sinx=cosx"; "since $x > 0$x>0, $\sqrt{x} = x^{1/2}$x​=x1/2 is well-defined". Phrases that lose marks: "differentiating $\dfrac{2}{x}$x2​ gives $\dfrac{2}{1} = 2$12​=2" (treating $x$x in the denominator as if it were a constant); leaving $x^{-1/2}$x−1/2 in the final answer when the question asks for surd form; differentiating $\sin x$sinx to $-\cos x$−cosx (sign error from confusing it with $\cos x$cosx).

A specific Edexcel pattern to watch: questions phrased "leaving each term in its simplest form" or "in surd form" are binding instructions on presentation. An answer that is mathematically correct but presented as $3x^{-3/2}$3x−3/2 when surd form was requested loses the final A1.

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Grade-band model answers

3-mark question

Question: Differentiate $y = 4x^3 + \dfrac{1}{x^2} - 2\cos x$y=4x3+x21​−2cosx with respect to $x$x.

Grade C response (~210 words):

Rewrite $\dfrac{1}{x^2}$x21​ as $x^{-2}$x−2.

Then $y = 4x^3 + x^{-2} - 2\cos x$y=4x3+x−2−2cosx.

Differentiate term-by-term:

$\dfrac{d}{dx}(4x^3) = 12x^2$dxd​(4x3)=12x2

$\dfrac{d}{dx}(x^{-2}) = -2x^{-3}$dxd​(x−2)=−2x−3

$\dfrac{d}{dx}(-2\cos x) = 2\sin x$dxd​(−2cosx)=2sinx (because $\dfrac{d}{dx}\cos x = -\sin x$dxd​cosx=−sinx, and the leading minus flips it back).

So $\dfrac{dy}{dx} = 12x^2 - 2x^{-3} + 2\sin x$dxdy​=12x2−2x−3+2sinx.