Unit 24 Exam Practice
Differentiation Techniques
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
SL 1 Paper 1
The functions \(f\) and \(g\) are differentiable for all \(x \in \mathbb{R}\). The table shows some values of \(f\), \(g\) and their derivatives.
| \(x\) | \(f(x)\) | \(f'(x)\) | \(g(x)\) | \(g'(x)\) |
|---|---|---|---|---|
| \(1\) | \(2\) | \(4\) | \(3\) | \(-2\) |
| \(3\) | \(5\) | \(-1\) | \(1\) | \(6\) |
(a) Let \(h(x) = f(x)\,g(x)\). Find \(h'(1)\). [2]
(b) Let \(k(x) = f\big(g(x)\big)\). Find \(k'(1)\). [2]
(c) Let \(m(x) = \dfrac{g(x)}{f(x)}\). Find \(m'(3)\). [2]
(d) Let \(n(x) = \ln\big(f(g(x))\big)\). Find \(n'(1)\). [2]
Total 8 marks
(a)
\(h'(1) = f'(1)\,g(1) + f(1)\,g'(1)\) M1
\(= 4 \times 3 + 2 \times (-2) = 8\) A1
[2 marks]
(b)
\(k'(1) = f'\big(g(1)\big) \times g'(1) = f'(3) \times g'(1)\) M1
\(= (-1) \times (-2) = 2\) A1
[2 marks]
(c)
\(m'(3) = \dfrac{f(3)\,g'(3) - g(3)\,f'(3)}{\big(f(3)\big)^{2}}\) M1
\(= \dfrac{5 \times 6 - 1 \times (-1)}{25} = \dfrac{31}{25}\) A1
[2 marks]
(d)
\(n'(x) = \dfrac{f'\big(g(x)\big)\,g'(x)}{f\big(g(x)\big)}\) M1
\(n'(1) = \dfrac{(-1) \times (-2)}{f(3)} = \dfrac{2}{5}\) A1
[2 marks]
SL 2 Paper 1
Let \(f(x) = x\sqrt{a - x^{2}}\), for \(-\sqrt{a} < x < \sqrt{a}\), where \(a > 1\) is a constant.
(a) Show that \(f'(x) = \dfrac{a - 2x^{2}}{\sqrt{a - x^{2}}}\). [4]
(b) Given that \(f'(1) = \dfrac{3}{2}\), find the value of \(a\). [4]
Total 8 marks
(a)
\(f'(x) = \sqrt{a - x^{2}} + x \times \dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{a - x^{2}}\) (product rule) M1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(a - x^{2}\right)^{\frac{1}{2}} = \dfrac{1}{2}\left(a - x^{2}\right)^{-\frac{1}{2}}(-2x) = -\dfrac{x}{\sqrt{a - x^{2}}}\) A1
\(f'(x) = \sqrt{a - x^{2}} - \dfrac{x^{2}}{\sqrt{a - x^{2}}}\) A1
\(= \dfrac{\left(a - x^{2}\right) - x^{2}}{\sqrt{a - x^{2}}} = \dfrac{a - 2x^{2}}{\sqrt{a - x^{2}}}\) A1 AG
[4 marks]
(b)
\(\dfrac{a - 2}{\sqrt{a - 1}} = \dfrac{3}{2}\) M1
\(4(a - 2)^{2} = 9(a - 1) \;\Rightarrow\; 4a^{2} - 25a + 25 = 0\) A1
\((4a - 5)(a - 5) = 0\), so \(a = 5\) or \(a = \dfrac{5}{4}\) A1
\(a = \dfrac{5}{4}\) gives \(f'(1) = \dfrac{-\frac{3}{4}}{\frac{1}{2}} = -\dfrac{3}{2}\), so it is rejected; \(a = 5\) R1
Squaring can give a false solution. Do not award the R1 if \(a = \dfrac{5}{4}\) is not rejected, or is rejected with no reason.
[4 marks]
See 24.3.
SL 3 Paper 2
Let \(f(x) = \sqrt{x}\,\ln x\), for \(x > 0\). The points \(P\) and \(Q\) on the graph of \(f\) have \(x\)-coordinates \(4\) and \(4 + h\), where \(h > 0\).
(a) Show that \(f'(x) = \dfrac{\ln x + 2}{2\sqrt{x}}\). [2]
(b) Find the gradient of the chord \(PQ\) when \(h = 0.1\) and when \(h = 0.01\). [2]
(c) Find the exact value of \(f'(4)\) in the form \(\dfrac{1 + \ln a}{b}\), where \(a, b \in \mathbb{Z}^{+}\). Explain how this value is related to your answers to part (b). [2]
(d) Find the two values of \(x\) for which the gradient of the graph of \(f\) is \(\dfrac{1}{2}\). [2]
Total 8 marks
(a)
\(f'(x) = \dfrac{1}{2\sqrt{x}}\ln x + \sqrt{x} \times \dfrac{1}{x}\) (product rule) M1
\(= \dfrac{\ln x}{2\sqrt{x}} + \dfrac{2}{2\sqrt{x}} = \dfrac{\ln x + 2}{2\sqrt{x}}\) A1 AG
[2 marks]
(b)
\(h = 0.1\): \(\dfrac{f(4.1) - f(4)}{0.1} = 0.844\) \((0.844421\ldots)\) A1
\(h = 0.01\): \(\dfrac{f(4.01) - f(4)}{0.01} = 0.846\) \((0.846357\ldots)\) A1
[2 marks]
(c)
\(f'(4) = \dfrac{\ln 4 + 2}{4} = \dfrac{2\ln 2 + 2}{4} = \dfrac{1 + \ln 2}{2}\) A1
\(f'(4) = 0.847\) \((0.846573\ldots)\): as \(h \to 0\), the gradient of the chord \(PQ\) approaches \(f'(4)\), the gradient of the curve at \(P\) R1
[2 marks]
(d)
\(\dfrac{\ln x + 2}{2\sqrt{x}} = \dfrac{1}{2}\) (for example, using the intersection of the graphs of \(y = f'(x)\) and \(y = 0.5\) on a GDC) M1
\(x = 0.215\) \((0.215223\ldots)\) or \(x = 28.7\) \((28.6941\ldots)\) A1
Award the A1 only if both values are given.
[2 marks]
SL 4 Paper 1
Let \(f(x) = \dfrac{\ln x}{x^{2}}\), for \(x > 0\).
(a) Show that \(f'(x) = \dfrac{1 - 2\ln x}{x^{3}}\). [3]
(b) Find the exact value of \(x\) for which \(f'(x) = 0\). [2]
(c) Show that \(f''(x) = \dfrac{6\ln x - 5}{x^{4}}\). [4]
(d) Find the exact value of \(f''(x)\) at the value of \(x\) found in part (b). [2]
Let \(g(x) = f\left(\mathrm{e}^{x}\right)\), for \(x \in \mathbb{R}\).
(e) Use the chain rule and part (a) to show that \(g'(x) = (1 - 2x)\,\mathrm{e}^{-2x}\). [3]
Total 14 marks
(a)
\(f'(x) = \dfrac{x^{2} \times \frac{1}{x} - \ln x \times 2x}{x^{4}}\) (quotient rule) M1 A1
\(= \dfrac{x - 2x\ln x}{x^{4}} = \dfrac{1 - 2\ln x}{x^{3}}\) A1 AG
[3 marks]
(b)
\(1 - 2\ln x = 0 \;\Rightarrow\; \ln x = \dfrac{1}{2}\) M1
\(x = \sqrt{\mathrm{e}}\) \(\left(= \mathrm{e}^{\frac{1}{2}}\right)\) A1
[2 marks]
(c)
\(f''(x) = \dfrac{x^{3} \times \left(-\frac{2}{x}\right) - (1 - 2\ln x) \times 3x^{2}}{x^{6}}\) M1 A1 A1
\(= \dfrac{-2x^{2} - 3x^{2} + 6x^{2}\ln x}{x^{6}}\) A1
\(= \dfrac{6\ln x - 5}{x^{4}}\) AG
Award M1 for the quotient rule, A1 for the derivative \(-\dfrac{2}{x}\) of the numerator, and A1 for the derivative \(3x^{2}\) of the denominator.
[4 marks]
(d)
\(f''\left(\sqrt{\mathrm{e}}\right) = \dfrac{6 \times \frac{1}{2} - 5}{\left(\sqrt{\mathrm{e}}\right)^{4}}\) M1
\(= -\dfrac{2}{\mathrm{e}^{2}}\) A1
[2 marks]
(e)
\(g'(x) = f'\left(\mathrm{e}^{x}\right) \times \mathrm{e}^{x}\) M1
\(= \dfrac{1 - 2\ln \mathrm{e}^{x}}{\left(\mathrm{e}^{x}\right)^{3}} \times \mathrm{e}^{x} = \dfrac{1 - 2x}{\mathrm{e}^{3x}} \times \mathrm{e}^{x}\) A1
\(= \dfrac{1 - 2x}{\mathrm{e}^{2x}} = (1 - 2x)\,\mathrm{e}^{-2x}\) A1 AG
[3 marks]
SL 5 Paper 1
Let \(f(x) = \ln(2 + \sin x)\), for \(0 \leq x \leq 2\pi\).
(a) Find \(f'(x)\). [2]
(b) Find the exact gradient of the graph of \(f\) at the point where \(x = \dfrac{\pi}{6}\). [2]
(c) Show that \(f''(x) = -\dfrac{1 + 2\sin x}{(2 + \sin x)^{2}}\). [4]
(d) Solve the equation \(f''(x) = 0\). [3]
(e) Show that \(f''(x) + \big(f'(x)\big)^{2} = -\dfrac{\sin x}{2 + \sin x}\). [3]
Total 14 marks
(a)
\(f'(x) = \dfrac{1}{2 + \sin x} \times \cos x\) (chain rule) M1
\(= \dfrac{\cos x}{2 + \sin x}\) A1
[2 marks]
(b)
\(f'\left(\dfrac{\pi}{6}\right) = \dfrac{\frac{\sqrt{3}}{2}}{2 + \frac{1}{2}}\) M1
\(= \dfrac{\sqrt{3}}{5}\) A1
[2 marks]
(c)
\(f''(x) = \dfrac{(2 + \sin x)(-\sin x) - \cos x \times \cos x}{(2 + \sin x)^{2}}\) (quotient rule) M1 A1
\(= \dfrac{-2\sin x - \left(\sin^{2} x + \cos^{2} x\right)}{(2 + \sin x)^{2}}\) A1
\(= \dfrac{-2\sin x - 1}{(2 + \sin x)^{2}} = -\dfrac{1 + 2\sin x}{(2 + \sin x)^{2}}\) A1 AG
[4 marks]
(d)
\(1 + 2\sin x = 0 \;\Rightarrow\; \sin x = -\dfrac{1}{2}\) M1
\(x = \dfrac{7\pi}{6}\) or \(x = \dfrac{11\pi}{6}\) A1 A1
[3 marks]
(e)
\(f''(x) + \big(f'(x)\big)^{2} = \dfrac{-1 - 2\sin x + \cos^{2} x}{(2 + \sin x)^{2}}\) M1
\(= \dfrac{-2\sin x - \sin^{2} x}{(2 + \sin x)^{2}}\) (using \(\cos^{2} x = 1 - \sin^{2} x\)) A1
\(= \dfrac{-\sin x\,(2 + \sin x)}{(2 + \sin x)^{2}} = -\dfrac{\sin x}{2 + \sin x}\) A1 AG
[3 marks]
HL questions
HL questions on this unit also use the AHL content: differentiation from first principles, the derivatives of the reciprocal and inverse trigonometric functions, of \(a^{x}\) and of \(\log_a x\), implicit differentiation, and higher derivatives (with a proof by mathematical induction from Unit 6), together with the SL content.
HL 1 Paper 1
Let \(f(x) = \dfrac{1}{x^{2} + 1}\), for \(x \in \mathbb{R}\).
(a) Use differentiation from first principles to show that \(f'(x) = -\dfrac{2x}{\left(x^{2} + 1\right)^{2}}\). [5]
(b) Hence find the gradient of the graph of \(y = \dfrac{5}{x^{2} + 1}\) at the point where \(x = 2\). [2]
Total 7 marks
(a)
\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{1}{h}\left(\dfrac{1}{(x + h)^{2} + 1} - \dfrac{1}{x^{2} + 1}\right)\) M1
\(= \dfrac{\left(x^{2} + 1\right) - \left((x + h)^{2} + 1\right)}{h\left((x + h)^{2} + 1\right)\left(x^{2} + 1\right)}\) A1
\(= \dfrac{-2xh - h^{2}}{h\left((x + h)^{2} + 1\right)\left(x^{2} + 1\right)}\) A1
\(= \dfrac{-2x - h}{\left((x + h)^{2} + 1\right)\left(x^{2} + 1\right)}\), for \(h \neq 0\) A1
letting \(h \to 0\): \(f'(x) = \dfrac{-2x - 0}{\left(x^{2} + 1\right)\left(x^{2} + 1\right)} = -\dfrac{2x}{\left(x^{2} + 1\right)^{2}}\) A1 AG
Do not award the final A1 if \(h\) is set equal to \(0\) before \(h\) has been cancelled.
[5 marks]
(b)
\(y = 5f(x)\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 5f'(x) = -\dfrac{10x}{\left(x^{2} + 1\right)^{2}}\) M1
at \(x = 2\): \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{20}{25} = -\dfrac{4}{5}\) A1
[2 marks]
See 24.2.
HL 2 Paper 1
Let \(f(x) = \arccos\left(1 - 2x^{2}\right)\), for \(-1 \leq x \leq 1\).
(a) Show that \(f'(x) = \dfrac{2}{\sqrt{1 - x^{2}}}\) for \(0 < x < 1\). [4]
(b) Find an expression for \(f'(x)\) for \(-1 < x < 0\). [2]
Total 6 marks
(a)
\(f'(x) = -\dfrac{1}{\sqrt{1 - \left(1 - 2x^{2}\right)^{2}}} \times (-4x)\) (chain rule) M1 A1
\(1 - \left(1 - 2x^{2}\right)^{2} = 4x^{2} - 4x^{4} = 4x^{2}\left(1 - x^{2}\right)\) A1
\(0 < x < 1\), so \(\sqrt{4x^{2}\left(1 - x^{2}\right)} = 2x\sqrt{1 - x^{2}}\) R1
\(f'(x) = \dfrac{4x}{2x\sqrt{1 - x^{2}}} = \dfrac{2}{\sqrt{1 - x^{2}}}\) AG
[4 marks]
(b)
for \(-1 < x < 0\): \(\sqrt{4x^{2}} = 2|x| = -2x\) M1
\(f'(x) = \dfrac{4x}{-2x\sqrt{1 - x^{2}}} = -\dfrac{2}{\sqrt{1 - x^{2}}}\) A1
[2 marks]
See 24.4.
HL 3 Paper 2
Consider the curves \(y = 2^{x}\) and \(y = \log_2 x\), for \(x > 0\).
(a) Write down \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) for each curve. [2]
(b) The gradient of the curve \(y = 2^{x}\) at the point \(A\) is \(1\). Find the exact coordinates of \(A\). [3]
(c) Find the value of \(x\) at which the two curves have the same gradient. [3]
Total 8 marks
(a)
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2^{x}\right) = 2^{x}\ln 2\) A1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\log_2 x\right) = \dfrac{1}{x\ln 2}\) A1
[2 marks]
(b)
\(2^{x}\ln 2 = 1\) M1
\(y = 2^{x} = \dfrac{1}{\ln 2}\) A1
\(x = \log_2\left(\dfrac{1}{\ln 2}\right) = -\log_2(\ln 2)\), so \(A\left(-\log_2(\ln 2),\ \dfrac{1}{\ln 2}\right)\) A1
Accept equivalent exact forms, for example \(x = -\dfrac{\ln(\ln 2)}{\ln 2}\).
[3 marks]
(c)
\(2^{x}\ln 2 = \dfrac{1}{x\ln 2}\) M1
\(x\,2^{x}(\ln 2)^{2} = 1\) (for example, using the intersection of the two graphs on a GDC) A1
\(x = 1.02\) \((1.02371\ldots)\) A1
[3 marks]
See 24.5.
HL 4 Paper 1
The curve \(C\) has equation \(xy + y^{2} = x^{3} + 1\).
(a) Find the \(y\)-coordinates of the two points on \(C\) where \(x = 1\). [2]
(b) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^{2} - y}{x + 2y}\). [2]
(c) Find the value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) at the point on \(C\) where \(x = 1\) and \(y < 0\). [4]
Total 8 marks
(a)
\(y + y^{2} = 1 + 1 \;\Rightarrow\; y^{2} + y - 2 = 0\) M1
\((y + 2)(y - 1) = 0\), so \(y = 1\) or \(y = -2\) A1
[2 marks]
(b)
\(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}\) M1 A1
\((x + 2y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - y\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^{2} - y}{x + 2y}\) AG
Award M1 for differentiating implicitly, with the product rule for \(xy\).
[2 marks]
(c)
at \((1,\,-2)\): \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3 + 2}{1 - 4} = -\dfrac{5}{3}\) A1
differentiating \(y + (x + 2y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}\) again with respect to \(x\) M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} + \left(1 + 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} + (x + 2y)\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6x\) A1
\(-\dfrac{5}{3} + \dfrac{35}{9} - 3\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6\), so \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\dfrac{34}{27}\) A1
Accept the quotient rule applied to the result of part (b).
[4 marks]
HL 5 Paper 1
Let \(y = \sec x + \tan x\), for \(-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\).
(a) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = y\sec x\). [2]
(b) Hence show that \(\cos x\,\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = y^{2}\). [4]
(c) Find the exact value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) when \(x = \dfrac{\pi}{3}\), giving your answer in the form \(p + q\sqrt{3}\), where \(p, q \in \mathbb{Z}\). [2]
Total 8 marks
(a)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sec x\tan x + \sec^{2} x\) A1 A1
\(= \sec x\,(\tan x + \sec x) = y\sec x\) AG
[2 marks]
(b)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{\mathrm{d}y}{\mathrm{d}x}\sec x + y\sec x\tan x\) (product rule) M1 A1
\(= y\sec^{2} x + y\sec x\tan x\) A1
\(= y\sec x\,(\sec x + \tan x) = y^{2}\sec x\) A1
so \(\cos x\,\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = y^{2}\) AG
[4 marks]
(c)
\(x = \dfrac{\pi}{3}\): \(\sec x = 2\), \(\tan x = \sqrt{3}\), so \(y = 2 + \sqrt{3}\) A1
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\left(2 + \sqrt{3}\right)^{2} = 14 + 8\sqrt{3}\) A1
[2 marks]
HL 6 Paper 1
Let \(f(x) = x\,3^{x}\), for \(x \in \mathbb{R}\).
(a) Show that \(f'(x) = 3^{x}(1 + x\ln 3)\). [2]
(b) Show that \(f''(x) = (\ln 3)\,3^{x}(2 + x\ln 3)\). [3]
(c) Prove by mathematical induction that \(f^{(n)}(x) = (\ln 3)^{n-1}\,3^{x}(n + x\ln 3)\) for all \(n \in \mathbb{Z}^{+}\). [6]
(d) Hence solve the equation \(f^{(10)}(x) = 0\). [2]
(e) Show that \(f^{(n+1)}(x) - (\ln 3)\,f^{(n)}(x) = (\ln 3)^{n}\,3^{x}\) for all \(n \in \mathbb{Z}^{+}\). [2]
Total 15 marks
(a)
\(f'(x) = 1 \times 3^{x} + x \times 3^{x}\ln 3\) (product rule) M1 A1
\(= 3^{x}(1 + x\ln 3)\) AG
[2 marks]
(b)
\(f''(x) = 3^{x}\ln 3\,(1 + x\ln 3) + 3^{x}\ln 3\) M1 A1
\(= (\ln 3)\,3^{x}(2 + x\ln 3)\) A1 AG
[3 marks]
(c)
let \(P(n)\) be the statement \(f^{(n)}(x) = (\ln 3)^{n-1}\,3^{x}(n + x\ln 3)\)
\(n = 1\): \((\ln 3)^{0}\,3^{x}(1 + x\ln 3) = 3^{x}(1 + x\ln 3) = f'(x)\), so \(P(1)\) is true A1
assume \(P(k)\) is true for some \(k \in \mathbb{Z}^{+}\): \(f^{(k)}(x) = (\ln 3)^{k-1}\,3^{x}(k + x\ln 3)\) M1
consider \(n = k + 1\): \(f^{(k+1)}(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}\left[(\ln 3)^{k-1}\,3^{x}(k + x\ln 3)\right]\) M1
\(= (\ln 3)^{k-1}\left[3^{x}\ln 3\,(k + x\ln 3) + 3^{x}\ln 3\right]\) (product rule) A1
\(= (\ln 3)^{k}\,3^{x}(k + 1 + x\ln 3)\), so \(P(k + 1)\) is true A1
since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^{+}\) R1
Award the final R1 only if at least four of the previous marks have been awarded. Do not award the second M1 if \(f^{(k+1)}(x)\) is written down from the formula with no differentiation.
[6 marks]
(d)
\(f^{(10)}(x) = (\ln 3)^{9}\,3^{x}(10 + x\ln 3)\), and \((\ln 3)^{9}\,3^{x} > 0\), so \(10 + x\ln 3 = 0\) M1
\(x = -\dfrac{10}{\ln 3}\) A1
[2 marks]
(e)
\(f^{(n+1)}(x) = (\ln 3)^{n}\,3^{x}(n + 1 + x\ln 3)\) (part (c) with \(n + 1\) in place of \(n\)) M1
\(f^{(n+1)}(x) - (\ln 3)\,f^{(n)}(x) = (\ln 3)^{n}\,3^{x}\big[(n + 1 + x\ln 3) - (n + x\ln 3)\big]\) A1
\(= (\ln 3)^{n}\,3^{x}\) AG
[2 marks]
HL 7 Paper 1
The curve \(C\) has equation \(\mathrm{e}^{x} + \mathrm{e}^{y} = \mathrm{e}^{x + y}\), for \(x > 0\).
(a) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{y - x}\). [5]
(b) Show that the equation of \(C\) can be written as \(y = x - \ln\left(\mathrm{e}^{x} - 1\right)\). [3]
(c) Use the equation in part (b) to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\), and show that your answer agrees with part (a). [3]
(d) Show that \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \mathrm{e}^{y - x}\left(1 + \mathrm{e}^{y - x}\right)\). [3]
(e) Find the exact values of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) at the point on \(C\) where \(x = \ln 3\). [2]
Total 16 marks
(a)
\(\mathrm{e}^{x} + \mathrm{e}^{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x + y}\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) M1 A1
\(\left(\mathrm{e}^{y} - \mathrm{e}^{x + y}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x + y} - \mathrm{e}^{x}\) A1
from the equation of \(C\): \(\mathrm{e}^{x + y} - \mathrm{e}^{x} = \mathrm{e}^{y}\) and \(\mathrm{e}^{y} - \mathrm{e}^{x + y} = -\mathrm{e}^{x}\) M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{e}^{y}}{-\mathrm{e}^{x}} = -\mathrm{e}^{y - x}\) A1 AG
[5 marks]
(b)
\(\mathrm{e}^{x + y} - \mathrm{e}^{y} = \mathrm{e}^{x}\) M1
\(\mathrm{e}^{y}\left(\mathrm{e}^{x} - 1\right) = \mathrm{e}^{x}\) A1
\(\mathrm{e}^{y} = \dfrac{\mathrm{e}^{x}}{\mathrm{e}^{x} - 1}\) (this is allowed, because \(\mathrm{e}^{x} - 1 > 0\) for \(x > 0\)) A1
\(y = \ln \mathrm{e}^{x} - \ln\left(\mathrm{e}^{x} - 1\right) = x - \ln\left(\mathrm{e}^{x} - 1\right)\) AG
[3 marks]
(c)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - \dfrac{\mathrm{e}^{x}}{\mathrm{e}^{x} - 1} = -\dfrac{1}{\mathrm{e}^{x} - 1}\) M1 A1
from (b), \(\mathrm{e}^{y - x} = \dfrac{\mathrm{e}^{y}}{\mathrm{e}^{x}} = \dfrac{1}{\mathrm{e}^{x} - 1}\)
so \(-\mathrm{e}^{y - x} = -\dfrac{1}{\mathrm{e}^{x} - 1}\): the two answers agree R1
[3 marks]
(d)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\mathrm{e}^{y - x}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} - 1\right)\) (chain rule) M1 A1
\(= -\mathrm{e}^{y - x}\left(-\mathrm{e}^{y - x} - 1\right)\) A1
\(= \mathrm{e}^{y - x}\left(1 + \mathrm{e}^{y - x}\right)\) AG
[3 marks]
(e)
\(x = \ln 3\): \(\mathrm{e}^{y - x} = \dfrac{1}{\mathrm{e}^{x} - 1} = \dfrac{1}{2}\) A1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{2}\), \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4}\) A1
[2 marks]
HL 8 Paper 1
Let \(f(x) = \dfrac{\arcsin x}{\sqrt{1 - x^{2}}}\), for \(-1 < x < 1\).
(a) By writing \(\sqrt{1 - x^{2}}\,f(x) = \arcsin x\), show that \(\left(1 - x^{2}\right)f'(x) - x\,f(x) = 1\). [3]
(b) Hence show that \(\left(1 - x^{2}\right)f''(x) - 3x\,f'(x) - f(x) = 0\). [3]
(c) Find the values of \(f(0)\), \(f'(0)\), \(f''(0)\) and \(f'''(0)\). [6]
You are given that, for all \(n \in \mathbb{Z}^{+}\), \[ \left(1 - x^{2}\right)f^{(n+1)}(x) - (2n + 1)\,x\,f^{(n)}(x) - n^{2} f^{(n-1)}(x) = 0, \] where \(f^{(0)}(x) = f(x)\).
(d) Find \(f^{(5)}(0)\), and explain why \(f^{(n)}(0) = 0\) for every positive even integer \(n\). [3]
Total 15 marks
(a)
\(\sqrt{1 - x^{2}}\,f'(x) + f(x)\,\dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{1 - x^{2}} = \dfrac{1}{\sqrt{1 - x^{2}}}\) (product rule) M1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{1 - x^{2}} = -\dfrac{x}{\sqrt{1 - x^{2}}}\) A1
\(\sqrt{1 - x^{2}}\,f'(x) - \dfrac{x\,f(x)}{\sqrt{1 - x^{2}}} = \dfrac{1}{\sqrt{1 - x^{2}}}\) A1
multiplying by \(\sqrt{1 - x^{2}}\): \(\left(1 - x^{2}\right)f'(x) - x\,f(x) = 1\) AG
[3 marks]
(b)
\(-2x\,f'(x) + \left(1 - x^{2}\right)f''(x) - f(x) - x\,f'(x) = 0\) M1 A1 A1
\(\left(1 - x^{2}\right)f''(x) - 3x\,f'(x) - f(x) = 0\) AG
Award M1 for differentiating both products, A1 for \(-2x\,f'(x) + \left(1 - x^{2}\right)f''(x)\) and A1 for \(-f(x) - x\,f'(x)\).
[3 marks]
(c)
\(f(0) = \arcsin 0 = 0\) A1
\(x = 0\) in (a): \(f'(0) - 0 = 1\), so \(f'(0) = 1\) A1
\(x = 0\) in (b): \(f''(0) - 0 - 0 = 0\), so \(f''(0) = 0\) A1
differentiating (b): \(\left(1 - x^{2}\right)f'''(x) - 5x\,f''(x) - 4f'(x) = 0\) M1 A1
\(x = 0\): \(f'''(0) - 4f'(0) = 0\), so \(f'''(0) = 4\) A1
[6 marks]
(d)
\(x = 0\) in the given result: \(f^{(n+1)}(0) = n^{2} f^{(n-1)}(0)\) M1
\(f^{(5)}(0) = 4^{2} f'''(0) = 16 \times 4 = 64\) A1
\(f^{(2m)}(0) = (2m - 1)^{2} f^{(2m-2)}(0)\), so every even derivative at \(0\) is a multiple of \(f(0) = 0\) R1
[3 marks]