Differentiation

• Rewrite as powers first; the power comes down and goes down by one; the chain rule multiplies by the derivative of the inside; the normal is \(-\dfrac{1}{m}\); and a stationary point's \(y\) comes from the curve, never from the derivative.
• Rewrite as powers, then \(nx^{n-1}\); outside, keep the inside, times the derivative of the inside; normal \(-\dfrac{1}{m}\); \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}\); \(\dfrac{dy}{dx} = 0\), \(y\) from the curve, and the second derivative for the nature.

• 1.7.1 I understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations f′(x), f″(x), dy/dx and d²y/dx² for first and second derivatives
• 1.7.2 I can use the derivative of xⁿ (for any rational n), together with constant multiples, sums and differences of functions, and of composite functions using the chain rule
• 1.7.3 I can apply differentiation to gradients, tangents and normals, increasing and decreasing functions and rates of change
• 1.7.4 I can locate stationary points and determine their nature, and use information about stationary points in sketching graphs

Accuracy for this chapter. Keep gradients exact where they are exact: \(\dfrac{3}{\sqrt{7}}\), \(-\dfrac{3}{4}\), \(-\dfrac{1}{8}\). Give rates and optimised values to 3 significant figures, with the unrounded value written first (\(12.566 \to 12.6\), \(0.063\,66 \to 0.0637\), \(5.4193 \to 5.42\), \(553.58 \to 554\)) and never round an intermediate value: carry \(r = 5.4193\) into \(S\), not \(5.42\). Near a minimum the two happen to agree to 3 s.f., because \(S\) is flat there; where a function is steep they need not. Every rate carries its units. Show the differentiation and the equation \(\dfrac{dy}{dx} = 0\) before its roots: a stationary point read off a calculator's table with no working earns nothing.

• “\(\dfrac{d}{dx}\left(\dfrac{1}{x^2}\right) = \dfrac{1}{2x}\) and \(\dfrac{d}{dx}(\sqrt{x}) = \tfrac{1}{2}\sqrt{x}\).” Correct Rewrite every root and reciprocal as a power before differentiating. \(\dfrac{1}{x^2} = x^{-2}\), so the derivative is \(-2x^{-3} = -\dfrac{2}{x^3}\). \(\sqrt{x} = x^{\frac{1}{2}}\), so the derivative is \(\tfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}\). The rule \(nx^{n-1}\) only works on \(x^n\); it has nothing to say about a fraction or a root until you have turned it into one.
• “\(\dfrac{d}{dx}\left((3x - 1)^{-2}\right) = -2(3x - 1)^{-3} \times (3x - 1)\).” Correct The chain rule multiplies by the derivative of the inside, not by the inside. The inside is \(3x - 1\) and its derivative is 3, so the answer is \(-2(3x - 1)^{-3} \times 3 = -\dfrac{6}{(3x - 1)^3}\). Forgetting the factor altogether, and writing \(-2(3x - 1)^{-3}\), is the other half of the same mistake.
• “The chord gradient is \(12 + 6h + h^2\), so the gradient of the curve at \(P\) is \(12 + 6h\).” Correct The gradient at \(P\) is the limit as \(h \to 0\): every term containing \(h\) vanishes, leaving 12.
• “The tangent has gradient 8, so the normal has gradient \(-8\).” Correct The normal is perpendicular to the tangent, so its gradient is \(-\dfrac{1}{m} = -\tfrac{1}{8}\): the negative reciprocal, not just the negative.
• “\(f\) is decreasing where \(f(x) < 0\).” Correct Decreasing is about the gradient: \(f'(x) < 0\). A function can be negative and increasing, or positive and decreasing. Give the answer as a set of values of \(x\), such as \(-1 < x < 4\).
• “\(\dfrac{dA}{dt} = 2\pi r = 8\pi\).” Correct \(2\pi r\) is \(\dfrac{dA}{dr}\), the rate with respect to the radius. The rate with respect to time needs the chain rule, written out before any number goes in: \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}\).
• “\(\dfrac{dy}{dx} = 0\) at \(x = 0\), so the stationary point is \((0, 0)\).” Correct \(\dfrac{dy}{dx} = 0\) gives the \(x\)-coordinate. The \(y\)-coordinate comes from substituting into the original equation of the curve: for \(y = x^3 - 3x^2 + 4\) the point is \((0, 4)\).
• “\(\dfrac{d^2y}{dx^2} = 0\), so the point is neither a maximum nor a minimum.” Correct Zero means the second-derivative test gives no information. Use the sign of \(\dfrac{dy}{dx}\) just either side instead: \(y = x^4\) has \(\dfrac{d^2y}{dx^2} = 0\) at the origin, and the origin is a minimum.
• “The gradient of the chord is \(12 + 6h + h^2\), so the gradient at \(P\) is \(12 + 6h\).” Repair The gradient at \(P\) is the limit as \(h \to 0\), which is 12; every term with \(h\) in it vanishes.
• “\(\dfrac{d}{dx}\left(\dfrac{1}{x^2}\right) = \dfrac{1}{2x}\).” Repair Write \(\dfrac{1}{x^2} = x^{-2}\) first; the derivative is \(-2x^{-3} = -\dfrac{2}{x^3}\).
• “\(\dfrac{d}{dx}(\sqrt{x}) = \tfrac{1}{2}\sqrt{x}\).” Repair \(\sqrt{x} = x^{\frac{1}{2}}\), so the derivative is \(\tfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}\); the power goes down by one, from \(\tfrac{1}{2}\) to \(-\tfrac{1}{2}\).
• “\(\dfrac{d}{dx}\left(\dfrac{3}{2x}\right) = \dfrac{3}{2}\).” Repair \(\dfrac{3}{2x} = \tfrac{3}{2}x^{-1}\), so the derivative is \(-\tfrac{3}{2}x^{-2} = -\dfrac{3}{2x^2}\). The \(x\) is in the denominator, so its power is negative.
• “\(\dfrac{d}{dx}\left(x^2(x - 3)\right) = 2x \times 1 = 2x\).” Repair A product is not differentiated factor by factor. Multiply out, \(x^3 - 3x^2\), then differentiate: \(3x^2 - 6x\).
• “\(\dfrac{d}{dx}\left(\sqrt{2x^3 + 5}\right) = \tfrac{1}{2}(2x^3 + 5)^{-\frac{1}{2}}\).” Repair Multiply by the derivative of the inside, \(6x^2\): the answer is \(\dfrac{3x^2}{\sqrt{2x^3 + 5}}\).
• “\(\dfrac{d}{dx}\left((3x - 1)^{-2}\right) = -2(3x - 1)^{-3} \times (3x - 1)\).” Repair The chain-rule factor is the derivative of the inside, 3, not the inside itself: \(-6(3x - 1)^{-3}\).
• “The normal at \((2, 1)\) has gradient \(-8\).” Repair The normal gradient is \(-\dfrac{1}{m} = -\tfrac{1}{8}\); \(-8\) is the negative of the tangent gradient, not its negative reciprocal.
• “The tangent at \(x = 2\) to \(y = x^2 - 3x\) is \(y = (2x - 3)x\).” Repair The tangent is a straight line with a number for its gradient. Substitute \(x = 2\) into \(\dfrac{dy}{dx}\) first, \(m = 1\), then write \(y + 2 = 1(x - 2)\).
• “\(f\) is decreasing where \(f(x) < 0\).” Repair Decreasing means \(f'(x) < 0\) — the sign of the derivative, not of the function.
• “\(12 - 3x^2 > 0\), so \(x < \pm 2\).” Repair \(x^2 < 4\) means \(-2 < x < 2\). “\(x < \pm 2\)” is not a set of values; sketch the quadratic and read off where it is positive.
• “\(\dfrac{dA}{dt} = 2\pi r = 8\pi\).” Repair \(2\pi r\) is \(\dfrac{dA}{dr}\); multiply by \(\dfrac{dr}{dt} = 0.5\) to get \(\dfrac{dA}{dt} = 4\pi\). Write the chain-rule line before substituting.
• “The volume is decreasing at 40, so \(\dfrac{dV}{dt} = 40\).” Repair A decreasing quantity has a negative rate: \(\dfrac{dV}{dt} = -40\). The answer can then be stated as “decreasing at…” with a positive number.
• “Stationary points: \(\dfrac{dy}{dx} = 0\) at \(x = 0\) and 2, so the points are \((0, 0)\) and \((2, 0)\).” Repair Substitute into the curve for \(y\): \((0, 4)\) and \((2, 0)\).
• “\(\dfrac{d^2y}{dx^2} = -6 < 0\), so \((0, 4)\) is a minimum.” Repair A negative second derivative means a maximum (\(\cap\), “sad”).
• “\(\dfrac{d^2y}{dx^2} = 0\) at the stationary point, so it is neither a maximum nor a minimum.” Repair The test is inconclusive, not negative: check the sign of \(\dfrac{dy}{dx}\) either side. \(y = x^4\) has a minimum at \(x = 0\) with \(\dfrac{d^2y}{dx^2} = 0\) there.
• Minimising \(S = 2\pi r^2 + 2\pi rh\) by differentiating with \(h\) still in it. Repair Eliminate \(h\) with the constraint \(\pi r^2 h = 1000\) first, so that \(S\) is a function of \(r\) alone.
• “\(r = 5.42\), so \(S = 2\pi(5.42)^2 + \dfrac{2000}{5.42} = 553.6\ \text{cm}^2\).” Repair Keep \(r\) unrounded (5.4193) through the substitution and round \(S\) once, to 3 significant figures: \(553.58 \to 554\ \text{cm}^2\).

• 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 8, differentiate every answer you integrate: it is this chapter run backwards. When you reach chapter 12, redo worked example 3; the chain rule there is the same rule applied to more functions, and the Paper 3 product and quotient rules sit beside it. When you reach chapter 16, redo worked example 6; a rate statement like “the volume increases at 20 cm3 per second” is how a differential equation is set up. When you reach chapter 19, redo the stationary-point card in section D with \(t\) for \(x\): velocity is the derivative of displacement, and its maximum is found the same way. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, 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 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. 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.
• Questions are structured, and this chapter's worked examples and mixed challenge combine the skills of section 1.7 in ways like these: differentiate (often by the chain rule), then find the tangent or normal at a point and where it meets an axis; find the stationary points, determine their nature, and sketch or use the curve; find the set of values for which a function is increasing or decreasing; a connected-rates problem with a formula supplied; and a practical maximum or minimum whose first part is “show that” the quantity can be written in one variable. The reasoning sits in the set-up lines: the rewritten powers, the chain-rule factor, the normal gradient, the equation \(\dfrac{dy}{dx} = 0\), and the elimination of the constraint.
• MF19 prints \(\dfrac{d}{dx}(x^n) = nx^{n-1}\), and the volume and surface area of a sphere and the volume of a cone for rates problems. The chain rule, the normal gradient, the tests for increasing and decreasing, connected rates and both stationary-point tests must be known. The rest of MF19's differentiation table is Paper 3. The MF19 card lists it all.
• Gradients that come out exact stay exact: \(\dfrac{3}{\sqrt{7}}\), \(-\dfrac{3}{4}\). Rates and optimised values are given to 3 significant figures with the unrounded value carried through, and every rate has units. A stationary point needs the derivative, the equation \(\dfrac{dy}{dx} = 0\) and its solution written down; coordinates read off a calculator's table without that working earn 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.