Edexcel A-Level Maths: Differentiation — Complete Revision Guide (9MA0)

Differentiation is the heart of Edexcel A-Level Maths (9MA0). It is the first half of the calculus content in the specification, sitting alongside integration as one of the two big technical pillars of the pure papers. Coordinate geometry depends on it for tangents and normals. Mechanics uses it constantly to move between displacement, velocity, and acceleration. Optimisation problems — finding the maximum volume of a box, or the minimum cost of a fence — are standard exam fare and rely on differentiation entirely. If your differentiation is fluent, a large fraction of the pure paper opens up. If it is shaky, the marks bleed away across topics that look unrelated on the surface.

This guide is a topic-by-topic walkthrough of the differentiation content in the 9MA0 specification. It covers everything Edexcel can examine in this section: differentiation from first principles, basic differentiation of standard functions, the chain rule, the product rule, the quotient rule, implicit differentiation, parametric differentiation, second derivatives, optimisation, and connected rates of change. For each topic you will see the core skills, the typical pitfalls, a short worked example or sketch, and a link to the full lesson on the LearningBro course.

The aim is not to replace working through problems. The only way to get good at differentiation is to do hundreds of derivatives until the rules become automatic. The aim is to give you a clear map of what you need to know, in the order Edexcel teaches it, so your revision is targeted rather than scattered. Use this guide as a checklist, a refresher, and a launchpad into focused practice.

What the Edexcel 9MA0 Specification Covers

The Edexcel A-Level Maths qualification (9MA0) is assessed through three two-hour papers, each worth 100 marks. Papers 1 and 2 cover pure mathematics, and Paper 3 covers statistics and mechanics. The differentiation content sits in Section 7 of the pure specification and can appear on either Paper 1 or Paper 2. There is no choice of questions and no coursework, so every mark must be earned in the exam.

Differentiation is one of the two highest-frequency topic areas on the pure papers, alongside algebra. It feeds directly into coordinate geometry questions about tangents and normals, into curve-sketching, into optimisation contexts, and into mechanics on Paper 3. The table below shows the sub-topics of Section 7, the part of the specification they sit under, and a realistic estimate of how many marks across a Paper 1 / Paper 2 sitting come from each.

Topic	Spec Section	Typical Paper 1/2 marks weight
Differentiation from first principles	7.1	3-5 marks
Basic differentiation of standard functions	7.2	4-6 marks
Chain rule	7.3	4-6 marks
Product rule	7.4	4-6 marks
Quotient rule	7.5	3-5 marks
Implicit differentiation	7.6	5-7 marks
Parametric differentiation	7.7	5-7 marks
Second derivatives	7.8	3-5 marks
Optimisation	7.9	6-10 marks
Connected rates of change	7.10	4-6 marks

These weights are estimates based on the spread of typical 9MA0 papers — not guarantees for any single year. What is reliable, however, is that differentiation is consistently one of the most heavily examined areas on Papers 1 and 2, and that the same skills resurface inside coordinate geometry and mechanics questions. Mastering this section is high-leverage revision.

Differentiation from First Principles

Differentiation from first principles is the formal definition of the derivative. The derivative of $f(x)$f(x) at a point is the limit of the gradient of a chord as the chord shrinks to zero length. Formally,

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$f′(x)=limh→0​hf(x+h)−f(x)​.

Edexcel expects you to be able to apply this definition directly to simple polynomial functions and, on a good year, to the trigonometric functions $\sin x$sinx and $\cos x$cosx (where the standard small-angle results $\sin h \approx h$sinh≈h and $\cos h \approx 1 - h^2 / 2$cosh≈1−h2/2 for small $h$h become useful).

The standard workflow is: write down $f(x + h) - f(x)$f(x+h)−f(x), expand carefully, divide by $h$h, simplify, and then take the limit as $h \to 0$h→0. The limit step usually consists of crossing out terms in $h$h that vanish. For $f(x) = x^2$f(x)=x2, the chord gradient is $((x + h)^2 - x^2) / h = (2xh + h^2) / h = 2x + h$((x+h)2−x2)/h=(2xh+h2)/h=2x+h, and the limit as $h \to 0$h→0 is $2x$2x. So $f'(x) = 2x$f′(x)=2x.

A common pitfall is dividing by $h$h before simplifying the numerator, which leads to messy fractions and lost track of cancelling terms. Another is sign errors when expanding $(x + h)^n$(x+h)n for small $n$n — write out each term carefully rather than relying on memorised binomial expansions. A third is taking the limit before cancelling — you cannot evaluate $h / h$h/h as $h \to 0$h→0 without first cancelling.

A short worked example. Differentiate $f(x) = x^3$f(x)=x3 from first principles. The numerator $f(x + h) - f(x) = (x + h)^3 - x^3 = 3x^2 h + 3x h^2 + h^3$f(x+h)−f(x)=(x+h)3−x3=3x2h+3xh2+h3. Dividing by $h$h gives $3x^2 + 3xh + h^2$3x2+3xh+h2, and as $h \to 0$h→0 the last two terms vanish, leaving $f'(x) = 3x^2$f′(x)=3x2.

For full coverage with practice questions and worked solutions, see the Differentiation from First Principles lesson.

Basic Differentiation

Once first principles has done its proof-of-concept work, you switch to standard differentiation rules. Edexcel expects you to know the derivatives of all the standard functions on the formula book and to apply linearity (the sum and constant-multiple rules) without writing them out.

The core derivatives you must have at your fingertips are:

Function $f(x)$f(x)	Derivative $f'(x)$f′(x)
$x^n$xn	$n x^{n-1}$nxn−1
$e^x$ex	$e^x$ex
$\ln x$lnx	$1/x$1/x
$\sin x$sinx	$\cos x$cosx
$\cos x$cosx	$-\sin x$−sinx
$\tan x$tanx	$\sec^2 x$sec2x
$a^x$ax	$a^x \ln a$axlna

The power rule $\frac{d}{dx}(x^n) = n x^{n-1}$dxd​(xn)=nxn−1 works for any real $n$n, including negative and fractional values. So $\frac{d}{dx}(1/x) = -1/x^2$dxd​(1/x)=−1/x2 and $\frac{d}{dx}(\sqrt{x}) = 1/(2\sqrt{x})$dxd​(x​)=1/(2x​). Linearity means $\frac{d}{dx}(a f(x) + b g(x)) = a f'(x) + b g'(x)$dxd​(af(x)+bg(x))=af′(x)+bg′(x) — constants stay outside the derivative.

A common pitfall is differentiating before simplifying. An expression like $f(x) = (x^2 + 3)/x$f(x)=(x2+3)/x should first be split into $x + 3/x = x + 3 x^{-1}$x+3/x=x+3x−1, then differentiated to $1 - 3 x^{-2} = 1 - 3/x^2$1−3x−2=1−3/x2. Trying to differentiate the original form requires the quotient rule and is a self-inflicted complication. Another pitfall is forgetting that $\frac{d}{dx}(\cos x) = -\sin x$dxd​(cosx)=−sinx has a minus sign — the cosine derivative is the most common sign-error source on the paper.

A short worked example. Differentiate $f(x) = 4x^3 - 5\sqrt{x} + 2/x^2 - e^x$f(x)=4x3−5x​+2/x2−ex. Rewrite as $4x^3 - 5x^{1/2} + 2x^{-2} - e^x$4x3−5x1/2+2x−2−ex. Differentiate term by term: $f'(x) = 12x^2 - (5/2) x^{-1/2} - 4 x^{-3} - e^x$f′(x)=12x2−(5/2)x−1/2−4x−3−ex.

For full coverage of every standard function and exam-style practice, see the Basic Differentiation lesson.

The Chain Rule

The chain rule differentiates composite functions: functions inside functions. If $y = f(g(x))$y=f(g(x)), then

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x).$dxdy​=f′(g(x))⋅g′(x).

In Leibniz notation, with $u = g(x)$u=g(x), $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$dxdy​=dudy​⋅dxdu​. The chain rule is the most heavily used differentiation rule at A-Level, because almost every interesting function in the syllabus is a composition.

The standard recipe is: identify the outer function and the inner function, differentiate the outer function leaving the inner function intact, then multiply by the derivative of the inner function. For $y = (3x + 2)^5$y=(3x+2)5, the outer function is "raise to the fifth", the inner is $3x + 2$3x+2. Differentiating gives $5(3x + 2)^4 \cdot 3 = 15(3x + 2)^4$5(3x+2)4⋅3=15(3x+2)4. The factor of 3 is the derivative of the inner — easy to forget under exam pressure.

For trigonometric chains like $y = \sin(2x + 1)$y=sin(2x+1), the same rule applies: $\frac{dy}{dx} = \cos(2x + 1) \cdot 2 = 2\cos(2x + 1)$dxdy​=cos(2x+1)⋅2=2cos(2x+1). For exponential chains like $y = e^{x^2}$y=ex2, $\frac{dy}{dx} = e^{x^2} \cdot 2x = 2x e^{x^2}$dxdy​=ex2⋅2x=2xex2. For logarithmic chains like $y = \ln(3x + 4)$y=ln(3x+4), $\frac{dy}{dx} = 1/(3x + 4) \cdot 3 = 3/(3x + 4)$dxdy​=1/(3x+4)⋅3=3/(3x+4).

A common pitfall is forgetting to multiply by the derivative of the inner function — writing $\frac{d}{dx} \sin(2x) = \cos(2x)$dxd​sin(2x)=cos(2x) instead of $2 \cos(2x)$2cos(2x). Another is mis-identifying the inner function for $y = (\sin x)^3$y=(sinx)3, which is $\sin x$sinx raised to the third (so the answer is $3 \sin^2 x \cdot \cos x$3sin2x⋅cosx), versus $y = \sin(x^3)$y=sin(x3), where the inner is $x^3$x3 (so the answer is $\cos(x^3) \cdot 3x^2$cos(x3)⋅3x2). The two look superficially similar in handwriting and the pairing is a classic exam trap.

For chain-rule drills with mixed function types, see the Chain Rule lesson.

The Product Rule

The product rule differentiates products of two functions. If $y = u v$y=uv, where $u = u(x)$u=u(x) and $v = v(x)$v=v(x), then

$\frac{dy}{dx} = u'v + u v'.$dxdy​=u′v+uv′.

A common verbal version is "first times derivative of second, plus second times derivative of first" — pick whichever order you prefer and stick to it, because mixing orders mid-problem is where most product-rule errors creep in.

The product rule applies whenever you have two functions of $x$x multiplied together, with neither one a constant. So $y = x^2 e^x$y=x2ex uses the product rule, but $y = 3 e^x$y=3ex does not (the 3 is a constant and stays outside the derivative). A clean workflow is to write out $u$u, $v$v, $u'$u′, $v'$v′ in a small table before applying the rule — this catches sign errors and stops you forgetting which derivative goes where.

A short worked example. Differentiate $y = x^2 \sin x$y=x2sinx. Set $u = x^2$u=x2, $v = \sin x$v=sinx, so $u' = 2x$u′=2x, $v' = \cos x$v′=cosx. Then $\frac{dy}{dx} = 2x \sin x + x^2 \cos x$dxdy​=2xsinx+x2cosx. You can factorise to $x(2 \sin x + x \cos x)$x(2sinx+xcosx) if a tidier form is needed.

A frequent pitfall is using the product rule when the chain rule would be simpler, or vice versa. For $y = (x + 1)^2$y=(x+1)2 you do not need the product rule — expand and differentiate term by term, or use the chain rule. The product rule is for genuinely distinct functions multiplied together. Another pitfall is forgetting one of the two terms in the formula — if the second function has a longer derivative, candidates sometimes write only the first term and stop.

For product-rule practice including problems where it combines with the chain rule, see the Product Rule lesson.

The Quotient Rule

The quotient rule differentiates a fraction of two functions. If $y = u / v$y=u/v, then

$\frac{dy}{dx} = \frac{u'v - u v'}{v^2}.$dxdy​=v2u′v−uv′​.

The order of the numerator matters — a sign flip would give the wrong answer — and the formula book reminder will save you only if you read it carefully. A common verbal mnemonic is "low D-high minus high D-low, all over low squared", where "high" is the numerator and "low" is the denominator. Whatever mnemonic you use, drill it until it is automatic.

The quotient rule applies whenever you have a fraction with $x$x in the denominator. For something like $y = (3x + 1)/(x - 2)$y=(3x+1)/(x−2), set $u = 3x + 1$u=3x+1, $v = x - 2$v=x−2, so $u' = 3$u′=3 and $v' = 1$v′=1. Then $\frac{dy}{dx} = (3(x - 2) - (3x + 1)(1))/(x - 2)^2 = (3x - 6 - 3x - 1)/(x - 2)^2 = -7/(x - 2)^2$dxdy​=(3(x−2)−(3x+1)(1))/(x−2)2=(3x−6−3x−1)/(x−2)2=−7/(x−2)2.

You can always avoid the quotient rule by rewriting $y = u / v$y=u/v as $y = u v^{-1}$y=uv−1 and using the product rule with the chain rule. Some candidates prefer this and never learn the quotient rule formula directly. Either approach is fine, but make a choice and be consistent. Switching between methods mid-paper is where errors creep in.

A common pitfall is reversing the numerator order, writing $u v' - u' v$uv′−u′v instead of $u' v - u v'$u′v−uv′. Another is forgetting to square the denominator. A third is failing to simplify — many quotient-rule answers can be tidied substantially by factoring out a common term in the numerator, and Edexcel mark schemes often want simplified form.

For quotient-rule practice with both direct and product-form approaches, see the Quotient Rule lesson.

Implicit Differentiation

So far, every function has been written explicitly: $y$y on one side, an expression in $x$x on the other. Some equations cannot be rearranged into that form, or rearranging is awkward. Implicit differentiation handles these. The trick is to differentiate both sides of the equation with respect to $x$x, treating $y$y as a function of $x$x and applying the chain rule whenever a $y$y-term is differentiated.

The single rule that powers everything is $\frac{d}{dx}(y^n) = n y^{n-1} \cdot \frac{dy}{dx}$dxd​(yn)=nyn−1⋅dxdy​, by the chain rule. So $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$dxd​(y2)=2ydxdy​, $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$dxd​(siny)=cosy⋅dxdy​, and so on. Each $y$y-derivative produces an extra $\frac{dy}{dx}$dxdy​ factor, which you then collect on one side of the equation and solve for.

A short worked example. Find $\frac{dy}{dx}$dxdy​ for the curve $x^2 + y^2 = 25$x2+y2=25 (a circle of radius 5). Differentiate both sides with respect to $x$x: $2x + 2y \frac{dy}{dx} = 0$2x+2ydxdy​=0. Solve for $\frac{dy}{dx}$dxdy​: $\frac{dy}{dx} = -x/y$dxdy​=−x/y. At the point $(3, 4)$(3,4) the gradient is $-3/4$−3/4.

For products like $xy$xy you need the product rule combined with implicit differentiation: $\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$dxd​(xy)=y+xdxdy​. So if the equation is $x^2 + xy + y^2 = 7$x2+xy+y2=7, differentiating gives $2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$2x+y+xdxdy​+2ydxdy​=0, so $\frac{dy}{dx} = -(2x + y)/(x + 2y)$dxdy​=−(2x+y)/(x+2y).

A common pitfall is forgetting the chain rule on $y$y-terms — writing $\frac{d}{dx}(y^2) = 2y$dxd​(y2)=2y instead of $2y \frac{dy}{dx}$2ydxdy​. Another is failing to apply the product rule to mixed terms like $xy$xy or $x \sin y$xsiny. A third is mis-collecting the $\frac{dy}{dx}$dxdy​ terms when there are several scattered through the expression.

For worked examples and tangent/normal questions on implicitly-defined curves, see the Implicit Differentiation lesson.

Parametric Differentiation

A parametric curve describes both $x$x and $y$y as functions of a third variable, called the parameter, often written as $t$t. So $x = f(t)$x=f(t) and $y = g(t)$y=g(t) jointly trace out a curve in the $xy$xy-plane. To find the gradient $\frac{dy}{dx}$dxdy​ on the curve, you do not need to eliminate $t$t. Instead you use

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$dxdy​=dx/dtdy/dt​.

This follows from the chain rule rearranged: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$dtdy​=dxdy​⋅dtdx​, so dividing through by $\frac{dx}{dt}$dtdx​ gives the result, provided $\frac{dx}{dt} \ne 0$dtdx​=0.

A short worked example. The curve $x = t^2 + 1$x=t2+1, $y = t^3 - 3t$y=t3−3t. Then $\frac{dx}{dt} = 2t$dtdx​=2t and $\frac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1)$dtdy​=3t2−3=3(t2−1). So $\frac{dy}{dx} = 3(t^2 - 1)/(2t)$dxdy​=3(t2−1)/(2t). To find the gradient at the point where $t = 2$t=2, substitute: gradient $= 3(4 - 1)/(4) = 9/4$=3(4−1)/(4)=9/4.

Edexcel often asks for the equation of a tangent or normal at a given parameter value, or for the points on the curve where $\frac{dy}{dx}$dxdy​ takes a specific value (often zero, for stationary points). The workflow is the same in each case: compute $\frac{dy}{dx}$dxdy​ in terms of $t$t, then handle the question. Stationary points come from $\frac{dy}{dx} = 0$dxdy​=0, which happens when the numerator $\frac{dy}{dt}$dtdy​ is zero (provided $\frac{dx}{dt} \ne 0$dtdx​=0 at the same value of $t$t).

A common pitfall is dividing the wrong way round, writing $\frac{dx/dt}{dy/dt}$dy/dtdx/dt​ instead of $\frac{dy/dt}{dx/dt}$dx/dtdy/dt​. Another is forgetting to substitute the parameter value back in to find the actual coordinates of the point — many tangent questions need both the gradient and the point. A third is missing the case where $\frac{dx}{dt} = 0$dtdx​=0 (a vertical tangent) and writing "no gradient" when the question expected a different observation.

For parametric tangent/normal practice and stationary-point problems, see the Parametric Differentiation lesson.

Second Derivatives

The second derivative $\frac{d^2 y}{dx^2}$dx2d2y​ (also written $f''(x)$f′′(x)) is the derivative of the derivative. It tells you how the gradient is changing — whether the curve is bending upwards or downwards. This information classifies stationary points and is the engine of optimisation problems.

At a stationary point where $\frac{dy}{dx} = 0$dxdy​=0:

Sign of $\frac{d^2 y}{dx^2}$dx2d2y​	Type of stationary point
Positive	Local minimum
Negative	Local maximum
Zero	Inconclusive — use the first-derivative sign test

The intuition is that a positive second derivative means the curve is concave up (like $y = x^2$y=x2), so a stationary point there must be a minimum. A negative second derivative means concave down, so the stationary point is a maximum. When the second derivative is zero, the test fails and you have to fall back on checking the sign of $\frac{dy}{dx}$dxdy​ either side of the stationary point.

A short worked example. Find and classify the stationary points of $y = x^3 - 3x + 2$y=x3−3x+2. Differentiating, $\frac{dy}{dx} = 3x^2 - 3$dxdy​=3x2−3, which is zero when $x^2 = 1$x2=1, so $x = 1$x=1 or $x = -1$x=−1. The second derivative is $\frac{d^2 y}{dx^2} = 6x$dx2d2y​=6x. At $x = 1$x=1, $6x = 6 > 0$6x=6>0, so $(1, 0)$(1,0) is a local minimum. At $x = -1$x=−1, $6x = -6 < 0$6x=−6<0, so $(-1, 4)$(−1,4) is a local maximum.

A common pitfall is using the first derivative to classify stationary points instead of the second. The first derivative is what you set to zero to find the stationary points; classifying them needs the second. Another pitfall is forgetting that "second derivative is zero" is inconclusive — students often assume it means a point of inflection, which is sometimes true but not always.

For stationary-point classification practice and points-of-inflection theory, see the Second Derivatives lesson.

Optimisation

Optimisation problems are word problems where you have to find the maximum or minimum of some quantity — a volume, an area, a cost, a time — subject to a constraint. They are the most heavily weighted single topic in the differentiation section and often appear as long-answer questions worth 8-12 marks.

The standard workflow has five stages. Stage 1: introduce variables for the unknowns. Stage 2: write down the quantity to be optimised in terms of those variables. Stage 3: use the constraint from the problem to eliminate all but one variable. Stage 4: differentiate, set the derivative to zero, and solve for the remaining variable. Stage 5: classify the stationary point with the second derivative and confirm it is the right type (max or min as the question asks). Without stage 5, full marks are not awarded.

A short worked example. A box with a square base and no lid is to be made from $300\,\text{cm}^2$300cm2 of card. Find the dimensions that maximise the volume. Let the base side be $x$x and the height be $h$h. The surface area is $x^2 + 4xh = 300$x2+4xh=300, so $h = (300 - x^2)/(4x)$h=(300−x2)/(4x). The volume is $V = x^2 h = x^2 (300 - x^2)/(4x) = (300x - x^3)/4 = 75x - x^3/4$V=x2h=x2(300−x2)/(4x)=(300x−x3)/4=75x−x3/4. Differentiating, $\frac{dV}{dx} = 75 - 3x^2 / 4$dxdV​=75−3x2/4, which is zero when $x^2 = 100$x2=100, so $x = 10$x=10 (taking the positive root). The second derivative is $-3x/2$−3x/2, which at $x = 10$x=10 is $-15 < 0$−15<0, so this is a maximum. The corresponding height is $h = (300 - 100)/(40) = 5$h=(300−100)/(40)=5, and the maximum volume is $V = 100 \times 5 = 500\,\text{cm}^3$V=100×5=500cm3.

A common pitfall is forgetting the constraint and trying to differentiate a two-variable expression. Another is solving for the wrong variable — if the question asks for the volume, you must compute it after finding $x$x, not stop at the value of $x$x. A third is skipping the second-derivative classification, which loses a method mark even when the answer is correct. A fourth is taking the negative root of a square when the context (a length) requires the positive root.

For optimisation practice on volumes, areas, and cost problems, see the Optimisation lesson.

Connected Rates of Change

Connected rates of change problems involve two or more quantities that vary with time and are linked by a geometric or algebraic relationship. The standard example is a balloon being inflated: the volume changes with time, the radius changes with time, and the volume formula links them. Edexcel asks you to find one rate given another.

The unifying tool is the chain rule applied to time. If $V$V depends on $r$r and $r$r depends on $t$t, then $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$dtdV​=drdV​⋅dtdr​. The unknown rate is whichever one you have not been told. The known relationship between the variables (the geometry of the problem) gives you $\frac{dV}{dr}$drdV​, and the question gives you one of the two time-derivatives.

A short worked example. A spherical balloon is inflated at $\frac{dV}{dt} = 50\,\text{cm}^3 \,\text{s}^{-1}$dtdV​=50cm3s−1. Find the rate at which the radius is increasing when $r = 5\,\text{cm}$r=5cm. The volume is $V = (4/3)\pi r^3$V=(4/3)πr3, so $\frac{dV}{dr} = 4\pi r^2$drdV​=4πr2. At $r = 5$r=5, $\frac{dV}{dr} = 100\pi$drdV​=100π. Then $\frac{dr}{dt} = \frac{dV/dt}{dV/dr} = 50/(100\pi) = 1/(2\pi) \,\text{cm s}^{-1}$dtdr​=dV/drdV/dt​=50/(100π)=1/(2π)cm s−1.

A common pitfall is dividing the wrong way round when applying the chain rule. Another is forgetting to substitute the specific value of the variable (here $r = 5$r=5) before computing the rate — the answer depends on the geometry at that particular moment.

For connected-rates problems including expanding spheres, draining cones, and shadow lengths, see the Connected Rates of Change lesson.

Common Mark-Loss Patterns Across Differentiation

Across the whole differentiation section, a small set of habits accounts for a disproportionate share of lost marks. None of these are about content you do not know. They are all about content you do know, applied carelessly.

• Forgetting the inner derivative in the chain rule. Writing $\frac{d}{dx} \sin(3x) = \cos(3x)$dxd​sin(3x)=cos(3x) instead of $3 \cos(3x)$3cos(3x) is the single most common chain-rule slip.
• Sign errors on the cosine derivative. $\frac{d}{dx}(\cos x) = -\sin x$dxd​(cosx)=−sinx. The minus sign is forgotten under exam pressure more often than any other.
• Reversed quotient rule. Writing $u v' - u' v$uv′−u′v instead of $u' v - u v'$u′v−uv′ flips every quotient-rule answer.
• Skipping the chain rule on $y$y-terms in implicit differentiation. Every $y$y-derivative produces an extra $\frac{dy}{dx}$dxdy​ factor — leaving it out wrecks the answer.
• Dividing the wrong way for parametric gradients. It is $\frac{dy/dt}{dx/dt}$dx/dtdy/dt​, not the other way round.
• Failing to classify stationary points. Setting $\frac{dy}{dx} = 0$dxdy​=0 is only half the work; the second-derivative check is the second half and earns a separate method mark.
• Forgetting the constraint in optimisation problems. You must reduce to one variable before differentiating; trying to differentiate a two-variable expression is a dead end.
• Stopping at $x$x when the question asks for the volume (or area, or whatever). Read the question again before you box your answer.
• Not simplifying. Many derivative answers can be tidied substantially. Mark schemes often want simplified form, and an unsimplified answer can lose an accuracy mark even when the working is correct.

Many candidates lose marks here every series. A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.

Recommended Six-Week Revision Plan

This plan is designed for a candidate who has covered the differentiation content in lessons but wants to revise it cleanly before the exam. It assumes about 5-6 hours per week on this section. Adjust pace if you are starting earlier or later.

Week	Topics	Practice
1	First principles; basic differentiation of standard functions	10 first-principles derivations; 30 basic-differentiation problems mixing polynomials, exponentials, logs, and trig
2	Chain rule; product rule	20 chain-rule problems; 15 product-rule problems including 5 that combine with chain rule
3	Quotient rule; implicit differentiation	15 quotient-rule problems; 15 implicit-differentiation problems including 5 with tangent/normal context
4	Parametric differentiation; second derivatives	15 parametric problems; 15 stationary-point classification problems
5	Optimisation; connected rates of change	8 full optimisation problems; 8 connected-rates problems
6	Mixed practice; targeted review of weakest topics; full differentiation question sets	One full mixed problem set per day; review marking-scheme working for any question scoring below 60%

The point of the plan is to keep moving forward while maintaining contact with earlier topics. Do not spend three weeks on the chain rule and run out of time before optimisation. By the end of week 5, every topic in the section should have had focused contact and a practice round. Week 6 is consolidation and weakness-targeting.

A useful discipline through the whole plan is to treat any question you got wrong not as a mistake but as a diagnostic. Was it a content gap? A method error? A careless arithmetic slip? Logging the cause means your next review session targets the right thing.

How LearningBro's Edexcel A-Level Maths Differentiation Course Helps

LearningBro's Edexcel A-Level Maths: Differentiation course is built around the structure of this guide. Each of the ten lessons covers one section of the 9MA0 specification, in the order Edexcel teaches it, with worked examples, practice questions and full mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.

The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each rule on the last. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling implicit differentiation for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.

If you want one place to revise this section of the spec well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the Edexcel A-Level Maths: Differentiation course.

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Related Reading

• Edexcel A-Level Maths: Algebra and Functions Revision Guide
• Edexcel A-Level Maths: Integration Revision Guide
• Edexcel A-Level Maths: Paper Strategy Guide