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Date	November 2021	Marks available	3	Reference code	21N.2.AHL.TZ0.8
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	8	Adapted from	N/A

Question

Consider the curve $C$ given by $y = x - x y \ln (x y)$ where $x > 0, y > 0$ .

Show that $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ .

[3]

Hence find the equation of the tangent to $C$ at the point where $x = 1$ .

[5]

Markscheme

METHOD 1

attempts to differentiate implicitly including at least one application of the product rule (M1)

$u = x y, v = \ln (x y), \frac{d u}{d x} = x \frac{d y}{d x} + y, \frac{d v}{d x} = (x \frac{d y}{d x} + y) \frac{1}{x y}$

$\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$ A1

Note: Award (M1)A1 for implicitly differentiating $y = x (1 - y \ln (x y))$ and obtaining $\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + x \frac{d y}{d x} \ln (x y) + y \ln (x y)]$ .

$\frac{d y}{d x} = 1 - [(x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y) (1 + \ln (x y))$ A1

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 2

$y = x - x y \ln x - x y \ln y$

attempts to differentiate implicitly including at least one application of the product rule (M1)

$\frac{d y}{d x} = 1 - (\frac{x y}{x} + (x \frac{d y}{d x} + y) \ln x) - (\frac{x y}{y} \frac{d y}{d x} + (x \frac{d y}{d x} + y) \ln y)$ A1

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - (x \ln x \frac{d y}{d x} + (1 + \ln x) y) - (y \ln y + x (\ln y \frac{d y}{d x} + \frac{d y}{d x}))$

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln x + \ln y + 1) - y (\ln x + \ln y + 1)$ A1

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln (x y) + 1) - y (\ln (x y) + 1)$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 3

attempt to differentiate implicitly including at least one application of the product rule M1

$u = x \ln (x y), v = y, \frac{d u}{d x} = \ln (x y) + (x \frac{d y}{d x} + y) \frac{x}{x y}, \frac{d v}{d x} = \frac{d y}{d x}$

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} \ln (x y) + y \ln (x y) + \frac{x y}{x y} (x \frac{d y}{d x} + y))$ A1

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln (x y) + 1) - y (\ln (x y) + 1)$ A1

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 4

lets $w = x y$ and attempts to find $\frac{d y}{d x}$ where $y = x - w \ln w$ M1

$\frac{d y}{d x} = 1 - (\frac{d w}{d x} + \frac{d w}{d x} \ln w) (= 1 - \frac{d w}{d x} (1 + \ln w))$ A1

$\frac{d w}{d x} = x \frac{d y}{d x} + y$ A1

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y + (x \frac{d y}{d x} + y) \ln (x y)) (= 1 - (x \frac{d y}{d x} + y) (1 + \ln (x y)))$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

[3 marks]

METHOD 1

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$ A1

substitutes $x = 1$ and their non-zero value of $y$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$2 \frac{d y}{d x} = 0 (\frac{d y}{d x} = 0)$ A1

equation of the tangent is $y = 1$ A1

METHOD 2

substitutes $x = 1$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (y)) = 1$

EITHER

correctly substitutes $\ln y = \frac{1 - y}{y}$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (1 + \frac{1}{y}) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

correctly substitutes $y + y \ln y = 1$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (2 + \ln y) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

THEN

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$

equation of the tangent is $y = 1$ A1

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal

Show 95 related questions

22M.3.AHL.TZ1.2c:
Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.
19M.1.SL.TZ2.S_9d:
The following diagram shows the graph of $f$ for 0 ≤ $x$ ≤ 3. Line $M$ is a tangent to the graph of $f$ at point P.

Given that $M$ is parallel to $L$ , find the $x$ -coordinate of P.
21N.1.SL.TZ0.5b:
Find $f (4)$ .
18N.2.SL.TZ0.T_4a:
Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.
18N.2.SL.TZ0.T_4b.iii:
Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.
18N.2.SL.TZ0.T_4c:
Sketch the graph of the function g (x) = 10x + 40 on the same axes.
19M.2.SL.TZ2.T_5b:
Write down the $y$ -intercept of the graph of $y = f\left( x \right)$ .
SPM.1.SL.TZ0.8b:
Find the value of $a$ and the value of $b$ .
19M.2.SL.TZ1.S_9b:
Find u.
19M.2.SL.TZ1.S_3b:
The graph of $f$ has a horizontal tangent line at $x = 0$ and at $x = a$ . Find $a$ .
17N.1.AHL.TZ0.H_11c:
Hence or otherwise, find an expression for the derivative of ${f_n}(x)$ with respect to $x$ .
17M.2.SL.TZ2.T_6f:
Write down the number of possible solutions to the equation $f(x) = 5$ .
19M.2.SL.TZ1.S_9d.iii:
Hence or otherwise, find the obtuse angle formed by the tangent line to $f$ at $x = 8$ and the tangent line to $f$ at $x = 2$ .
SPM.1.SL.TZ0.8c.i:
Sketch the graph of $y = f'\left( x \right)$ .
19M.1.AHL.TZ2.H_6b:
Hence find the equation of the normal to $C$ at the point (1, 1).
19N.1.SL.TZ0.S_10c:
The line ${L_2}$ is tangent to the graph of $g$ at ${\text{A}}$ and has equation $y = \left( {{\text{ln}}\,p} \right)x + q + 1$ .

The line ${L_2}$ passes through the point $\left( { - 2{\text{, }} - 2} \right)$ .

The gradient of the normal to $g$ at ${\text{A}}$ is $\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$ .

Find the equation of ${L_1}$ in terms of $x$ .
21N.1.SL.TZ0.5a:
Write down the value of $f' (4)$ .
17N.1.AHL.TZ0.H_11b:
By using mathematical induction, prove that

${f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}$ where $m \in \mathbb{Z}$ .
19N.2.AHL.TZ0.H_11a.i:
Using implicit differentiation, find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .
19N.1.SL.TZ0.S_10a:
Write down the coordinates of ${\text{B}}$ .
19M.2.SL.TZ1.S_9d.i:
Find $\left( {f \circ f} \right)\left( x \right)$ .
SPM.1.SL.TZ0.8a:
Find $f'\left( x \right)$ .
19N.2.SL.TZ0.S_3b:
Let ${\text{C}}$ be a point on the graph of $h$ . The tangent to the graph of $h$ at ${\text{C}}$ is parallel to the graph of $f$ .

Find the $x$ -coordinate of ${\text{C}}$ .
19M.2.SL.TZ2.T_5c:
Sketch the graph of $y = f\left( x \right)$ for −3 ≤ $x$ ≤ 3 and −4 ≤ $y$ ≤ 12.
19N.2.SL.TZ0.S_3a:
Find $h\left( x \right)$ .
18N.1.SL.TZ0.S_10b.i:
Find $f'\left( x \right)$ .
16N.1.SL.TZ0.T_14b:
Find the coordinates of P.
19N.2.AHL.TZ0.H_11a.ii:
Find the equation of the tangent to the curve at the point $\left( {\frac{1}{4}{\text{, }}\frac{{5\pi }}{6}} \right)$ .
19M.1.SL.TZ2.S_9a:
Find the value of ${\text{tan}}\,\theta$ .
18M.2.AHL.TZ1.H_9a:
Show that there are exactly two points on the curve where the gradient is zero.
18M.2.AHL.TZ1.H_9c:
The normal at P cuts the curve again at the point Q. Find the $x$ -coordinate of Q.
19M.2.AHL.TZ1.H_1:
Let $l$ be the tangent to the curve $y = x{{\text{e}}^{2x}}$ at the point (1, ${{\text{e}}^2}$ ).

Find the coordinates of the point where $l$ meets the $x$ -axis.
19M.1.SL.TZ2.S_9b:
Line $L$ passes through the origin and has a gradient of ${\text{tan}}\,\theta$ . Find the equation of $L$ .
22M.3.AHL.TZ1.2b:
Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .
22M.3.AHL.TZ1.2d.ii:
Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .
17M.2.AHL.TZ1.H_2a:
Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of $x$ and $y$ .
22M.1.SL.TZ2.7c:
Find the equation of $L$ .
SPM.1.SL.TZ0.8c.ii:
Hence explain why the graph of $f$ has a local maximum point at $x = a$ .
20N.1.SL.TZ0.T_13b:
Write down the gradient of this tangent.
SPM.1.SL.TZ0.8d.i:
Find $f''\left( b \right)$ .
19M.1.AHL.TZ2.H_6a:
At the point (1, 1) , show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 + \pi }}{{2 - \pi }}$ .
21M.1.SL.TZ1.5c:
Hence, show that $k = e + \frac{e^{2}}{4}$ .
21M.1.SL.TZ1.5a:
Find $f' (x)$ .
21M.1.SL.TZ1.5b:
Show that $h = \frac{e + 6}{2}$ .
19M.2.SL.TZ1.S_9d.ii:
Hence, write down ${f^{ - 1}}\left( x \right)$ .
21M.2.SL.TZ1.9b:
Find the exact coordinates of $Q$ .
20N.1.AHL.TZ0.H_11a:
Show that $\frac{d y}{d x} = \frac{y \cos (x y)}{2 y - x \cos (x y)}$ .
20N.1.SL.TZ0.S_10b:
Find the area of triangle $AOB$ in terms of $k$ .
18N.2.AHL.TZ0.H_5:
Differentiate from first principles the function $f\left( x \right) = 3{x^3} - x$ .
19M.2.SL.TZ1.S_9a:
Find the gradient of $L$ .
18M.2.AHL.TZ1.H_9b:
Find the equation of the normal to the curve at the point P.
20N.1.SL.TZ0.S_10c:
The graph of $f$ is translated by $(\begin{matrix} 4 \\ 3 \end{matrix})$ to give the graph of $g$ .
In the following diagram:
- point $Q$ lies on the graph of $g$
- points $C$ , $D$ and $E$ lie on the vertical asymptote of $g$
- points $D$ and $F$ lie on the horizontal asymptote of $g$
- point $G$ lies on the $x$ -axis such that $FG$ is parallel to $DC$ .
Line $L_{2}$ is the tangent to the graph of $g$ at $Q$ , and passes through $E$ and $F$ .

Given that triangle $EDF$ and rectangle $CDFG$ have equal areas, find the gradient of $L_{2}$ in terms of $p$ .
20N.1.AHL.TZ0.H_11b:
Prove that, when $\frac{d y}{d x} = 0, y = \pm 1$ .
20N.1.AHL.TZ0.H_2:
Find the equation of the tangent to the curve $y = e^{2 x} - 3 x$ at the point where $x = 0$ .
19N.1.SL.TZ0.S_10b:
Given that $f'\left( a \right) = \frac{1}{{{\text{ln}}\,p}}$ , find the equation of ${L_1}$ in terms of $x$ , $p$ and $q$ .
20N.1.SL.TZ0.S_10a.ii:
Show that the equation of $L_{1}$ is $k x + p^{2} y - 2 p k = 0$ .
EXN.2.AHL.TZ0.6b:
The tangent to $C$ at the point Ρ is parallel to the $y$ -axis.

Find the $x$ -coordinate of Ρ.
21M.1.SL.TZ2.5b:
Given that the gradient of $L$ is $\frac{1}{3}$ , find the $x$ -coordinate of $B$ .
SPM.1.SL.TZ0.8e:
The normal to the graph of $f$ at $x = a$ and the tangent to the graph of $f$ at $x = b$ intersect at the point ( $p$ , $q$ ) .

Find the value of $p$ and the value of $q$ .
20N.1.SL.TZ0.S_10a.i:
Find $f' (p)$ in terms of $k$ and $p$ .
17M.1.SL.TZ2.T_13b:
Find the equation of $N$ . Give your answer in the form $ax + by + d = 0$ where $a$ , $b$ , $d \in \mathbb{Z}$ .
21N.2.AHL.TZ0.8b:
Hence find the equation of the tangent to $C$ at the point where $x = 1$ .
22M.1.SL.TZ1.5:
Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in ℝ$ and $k \in ℚ$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Find the value of $k$ .
22M.3.AHL.TZ2.1f.i:
Find the equation of the tangent to $C$ at $P$ .
17M.2.SL.TZ2.T_6d.i:
Write down the $x$ -coordinates of these two points;
17M.1.SL.TZ2.T_13c:
Draw the line $N$ on the diagram above.
18M.1.SL.TZ1.T_5c:
Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.
17M.1.SL.TZ2.T_13a:
Write down the value of $f(1)$ .
17M.1.SL.TZ1.S_9c:
The line $y = kx - 5$ is a tangent to the curve of $f$ . Find the values of $k$ .
17N.1.AHL.TZ0.H_11a:
Determine whether ${f_n}$ is an odd or even function, justifying your answer.
21N.1.SL.TZ0.5c:
Find $h (4)$ .
SPM.1.SL.TZ0.8d.ii:
Hence, use your answer to part (d)(i) to show that the graph of $f$ has a local minimum point at $x = b$ .
19M.2.SL.TZ1.S_9c:
Find the acute angle between $y = x$ and $L$ .
17M.2.AHL.TZ1.H_2b:
Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$
20N.1.AHL.TZ0.H_11c:
Hence find the coordinates of all points on $C$ , for $0 < x < 4 π$ , where $\frac{d y}{d x} = 0$ .
16N.1.SL.TZ0.T_14a:
Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .
18M.2.AHL.TZ1.H_9d:
The shaded region is rotated by 2 $\pi$ about the $y$ -axis. Find the volume of the solid formed.
17N.1.AHL.TZ0.H_11d:
Show that, for $n > 1$ , the equation of the tangent to the curve $y = {f_n}(x)$ at $x = \frac{\pi }{4}$ is $4x - 2y - \pi = 0$ .
21M.2.SL.TZ1.9c:
Show that the equation of $L$ is $y = - x + 2 \ln 45 + 2$ .
EXN.2.AHL.TZ0.6a:
Show that $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ .
18M.1.SL.TZ1.T_5a:
Write down the coordinates of C, the midpoint of line segment AB.
18N.1.SL.TZ0.T_11c:
Find the equation of this tangent. Give your answer in the form y = mx + c.
20N.1.SL.TZ0.T_13c:
Find the value of $k$ .
18M.1.SL.TZ1.T_5b:
Find the gradient of the line DC.
17M.2.SL.TZ2.T_6a:
Write down the $y$ -intercept of the graph.
17M.2.SL.TZ2.T_6c.i:
Show that $a = 8$ .
17M.2.SL.TZ2.T_6b:
Find $f'(x)$ .
20N.1.SL.TZ0.T_13a:
Write down $f' (x)$ .
19M.2.SL.TZ2.T_5h:
Determine the range of $f\left( x \right)$ for $p$ ≤ $x$ ≤ $q$ .
17M.2.SL.TZ2.T_6g:
The equation $f(x) = m$ , where $m \in \mathbb{R}$ , has four solutions. Find the possible values of $m$ .
21N.1.SL.TZ0.5d:
Hence find the equation of the tangent to the graph of $h$ at $x = 4$ .
18M.2.SL.TZ1.T_4e:
Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.
17M.2.SL.TZ2.T_6c.ii:
Find $f(2)$ .
17M.2.SL.TZ2.T_6d.ii:
Write down the intervals where the gradient of the graph of $y = f(x)$ is positive.
17M.2.SL.TZ2.T_6e:
Write down the range of $f(x)$ .

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Topic 5 —Calculus » AHL 5.14—Implicit functions, related rates, optimisation

Topic 5 —Calculus

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