Differentiation (Pure 3)

• The applications never change: a gradient, then \(y - y_1 = m(x - x_1)\), \(-\dfrac1m\) for a normal, and zero for a stationary point. Only the source of the gradient is new.
• \(\ln f(x) \to \dfrac{f'(x)}{f(x)}\); \(\left(\dfrac uv\right)' = \dfrac{vu' - uv'}{v^2}\); \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\); \(\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}\). If these four come back instantly on day 30, the chapter has stuck.

• 3.4.1 2.4.1 I can use the derivatives of eˣ, ln x, sin x, cos x, tan x, tan⁻¹x, together with constant multiples, sums, differences and composites — with angles in radians, and with the tan⁻¹x part Paper 3 only
• 3.4.2 2.4.2 I can differentiate products and quotients
• 3.4.3 2.4.3 I can find and use the first derivative of a function which is defined parametrically or implicitly — the syllabus's note adds “including use in problems involving tangents and normals”, and this chapter also uses it for stationary points

Accuracy for this chapter. Leave exact answers exact: \(e^{-1/2}\), \(-\dfrac{1}{2e}\), \(\sqrt{\dfrac e2}\), \(\dfrac{\sqrt{21}}{3}\), \(\tfrac14\pi - \tfrac12\), \(-\tfrac12\ln 2\), \(\dfrac{16}{25}\). Give a decimal to 3 significant figures only when the question asks for one or after the exact form, and give \(x\) in radians (\(0.588\), never \(33.7^\circ\)). Keep unrounded values (\(0.588\,00\), \(1.5275\), \(3.0551\)) until the end, and show the equation you solved before any calculator value: an unsupported answer earns nothing.

• “\(\dfrac{d}{dx}\ln(2x + 1) = \dfrac{1}{2x + 1}\).” Correct The derivative of \(\ln f(x)\) is \(\dfrac{f'(x)}{f(x)}\): the derivative of the inside over the inside. \(\dfrac{d}{dx}\ln(2x + 1) = \dfrac{2}{2x + 1}\). The same rule gives \(\dfrac{d}{dx}\ln(5x) = \dfrac{5}{5x} = \dfrac1x\), not \(\dfrac5x\).
• “\(\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{u\dfrac{dv}{dx} - v\dfrac{du}{dx}}{v^2}\).” Correct Quotient rule: \(v\,u'\) minus \(u\,v'\), all over \(v^2\). Because of the minus sign, the other order gives the right size with the wrong sign. MF19 prints the rule in the correct order; copy it, with \(u, v, u', v'\) written down first.
• “Differentiating \(x^2 + y^2 = 25\) with respect to \(x\): \(2x + 2y = 0\).” Correct \(\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}\). \(y\) is a function of \(x\), so the chain rule attaches \(\dfrac{dy}{dx}\) to every term in \(y\): \(2x + 2y\dfrac{dy}{dx} = 0\). A constant differentiates to 0.
• “\(\dfrac{d}{dx}(\sin x) = \cos x\), so the gradient of \(y = \sin x\) at \(x = 30^\circ\) is \(\cos 30^\circ = 0.866\).” Correct \(\dfrac{d}{dx}(\sin x) = \cos x\) is true only when \(x\) is in radians. Work at \(x = \dfrac{\pi}{6}\): the gradient is \(\cos\dfrac{\pi}{6} = 0.866\) per radian. Every derivative, tangent and stationary point in Paper 3 is worked in radians.
• “\(x = t^2\), \(y = t^3 - 3t\) is stationary where \(\dfrac{dx}{dt} = 0\), at \(t = 0\).” Correct A parametric curve is stationary where \(\dfrac{dy}{dt} = 0\) (with \(\dfrac{dx}{dt} \ne 0\)). Where \(\dfrac{dx}{dt} = 0\) and \(\dfrac{dy}{dt} \ne 0\) the tangent is parallel to the \(y\)-axis, the opposite of stationary.
• “\(x^2 + y^2 = xy + 7\) is stationary where \(y = 2x\).” Correct \(y = 2x\) is a line, not a point. Substitute it into the curve's equation: \(3x^2 = 7\), giving the two points \(\left(\pm\dfrac{\sqrt{21}}{3}, \pm\dfrac{2\sqrt{21}}{3}\right)\).
• “\(\dfrac{d}{dx}\left(x^2\ln x\right) = 2x \times \dfrac1x = 2\).” Correct The derivative of a product is not the product of the derivatives. Product rule: \(2x\ln x + x^2 \cdot \dfrac1x = x(2\ln x + 1)\).
• “\(\dfrac{d}{dx}\ln(2x + 1) = \dfrac{1}{2x + 1}\).” Repair The derivative of the inside goes on top: \(\dfrac{2}{2x + 1}\).
• “\(\dfrac{d}{dx}\ln(5x) = \dfrac5x\).” Repair \(\dfrac{5}{5x} = \dfrac1x\). Equivalently \(\ln 5x = \ln 5 + \ln x\), and \(\ln 5\) is a constant.
• “\(\dfrac{d}{dx}e^{x^2} = x^2e^{x^2 - 1}\).” Repair \(e^x\) is not a power of \(x\), so the power rule does not apply. \(e^u \to u'e^u\): \(2xe^{x^2}\).
• “\(\dfrac{d}{dx}\cos x = \sin x\).” Repair \(-\sin x\). The MF19 table shows the minus.
• “\(\dfrac{d}{dx}\sin 3x = \cos 3x\).” Repair The chain rule gives \(3\cos 3x\): the derivative of the inside, 3, is a factor.
• Differentiating \(\sin x^\circ\) as \(\cos x^\circ\). Repair Calculus needs radians. In degrees the derivative carries a factor \(\dfrac{\pi}{180}\): \(\dfrac{d}{dx}(\sin x^\circ) = \dfrac{\pi}{180}\cos x^\circ\).
• “\(\dfrac{d}{dx}\left(x^2\ln x\right) = 2x \times \dfrac1x = 2\).” Repair The derivative of a product is not the product of the derivatives. Use \(v\dfrac{du}{dx} + u\dfrac{dv}{dx}\): \(2x\ln x + x\).
• “\(\dfrac{d}{dx}\left(\dfrac uv\right) = \dfrac{u'v'}{v^2}\)”, or “\(= \dfrac{u\dfrac{dv}{dx} - v\dfrac{du}{dx}}{v^2}\).” Repair \(\dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\), in that order; the other order gives the wrong sign.
• “Parametric: \(\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}\).” Repair \(y\) over \(x\): \(\dfrac{dy/dt}{dx/dt}\), as printed in MF19.
• “The curve is stationary where \(\dfrac{dx}{dt} = 0\).” Repair Stationary needs \(\dfrac{dy}{dt} = 0\) (with \(\dfrac{dx}{dt} \ne 0\)); \(\dfrac{dx}{dt} = 0\) gives a tangent parallel to the \(y\)-axis.
• “\(\dfrac{d}{dx}(y^2) = 2y\).” Repair \(y\) is a function of \(x\), so the chain rule gives \(2y\dfrac{dy}{dx}\).
• “\(\dfrac{d}{dx}(xy) = \dfrac{dy}{dx}\)”, or “\(= y\).” Repair Product rule: \(y + x\dfrac{dy}{dx}\).
• “\(\dfrac{d}{dx}(7) = 7\)” in an implicit equation. Repair Every constant differentiates to 0.
• Setting \(y = 2x\) for the stationary points of \(x^2 + y^2 = xy + 7\) and stopping there. Repair \(y = 2x\) is a line, not a point. Substitute it into the curve to find where the line meets it: \(3x^2 = 7\), \(x = \pm\dfrac{\sqrt{21}}{3}\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. When you reach chapter 13, re-answer retrieval questions 1 to 5 and read each answer backwards as an integral. When you reach chapter 14, re-sketch figure 3 from its derivative before locating a root. When you reach chapter 16, redo worked example 1: a differential equation is a derivative like this one, set equal to something. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A derivative asked for in one part and used in the next: “show that \(\dfrac{dy}{dx} = \ldots\)” followed by “hence find the stationary points”; a parametric or implicit curve with a tangent or normal at a named point; a stationary value combined with another section (an \(R\)-form, a logarithmic equation). The reasoning sits in the simplification: taking out a common factor so that \(\dfrac{dy}{dx} = 0\) can be solved, and substituting into the curve.
• The derivatives of \(e^x\), \(\ln x\), \(\sin x\), \(\cos x\), \(\tan x\) and \(\tan^{-1}x\), the product and quotient rules, and \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}\) are printed. The chain rule, the composite forms such as \(\dfrac{f'(x)}{f(x)}\), and implicit differentiation must be known. The MF19 card has the detail.
• Every angle is in radians. Exact coordinates stay exact (\(e^{-1/2}\), \(\dfrac{\sqrt{21}}{3}\), \(\tfrac14\pi - \tfrac12\)); decimals are to 3 significant figures. A “show that” derivative runs forwards from the given function. The equation you solved is written before any calculator value, because an unsupported answer earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.