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PSLE Foundation Mathematics

<p>The functions $\mathrm{p}$ and $\mathrm{q}$ are continuous for&nbsp;$1\leq x\leq5$. The function $\mathrm{p}$ is strictly increasing, while $\mathrm{q}$ is strictly decreasing and $\mathrm{q}(x)&gt;0$. Selected values are shown below.</p>
<div class="content-block content-block--table"><div class="table-container"><table class="table--original-style" style="table-layout: fixed; width: 300px;"><colgroup><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"><col style="width: 16.6667%;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center">$x$</th><th style="text-align: center; --cell-justify: center">$1$</th><th style="text-align: center; --cell-justify: center">$2$</th><th style="text-align: center; --cell-justify: center">$3$</th><th style="text-align: center; --cell-justify: center">$4$</th><th style="text-align: center; --cell-justify: center">$5$</th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">$\mathrm{p}(x)$</td><td style="text-align: center; --cell-justify: center">$-10$</td><td style="text-align: center; --cell-justify: center">$-8$</td><td style="text-align: center; --cell-justify: center">$-6$</td><td style="text-align: center; --cell-justify: center">$-4$</td><td style="text-align: center; --cell-justify: center">$5$</td></tr><tr><td style="text-align: center; --cell-justify: center">$\mathrm{q}(x)$</td><td style="text-align: center; --cell-justify: center">$5$</td><td style="text-align: center; --cell-justify: center">$4$</td><td style="text-align: center; --cell-justify: center">$3$</td><td style="text-align: center; --cell-justify: center">$2$</td><td style="text-align: center; --cell-justify: center">$1$</td></tr></tbody></table></div></div>
<p>Using the information in the table,</p>
<ol class="abc-list"><li data-list-item-id="q1-a" data-list-item-path="q1-a"><div class="list-item-paragraph"><p>Find $\mathrm{p}\mathrm{q}(2)$.</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="q1-b" data-list-item-path="q1-b"><div class="list-item-paragraph"><p>Solve $\mathrm{p}^{-1}\mathrm{q}(x)=5$.</p><span class="inline-marks">[2]</span></div></li></ol>
<p>The function $\mathrm{r}$ is defined by $\mathrm{r}(x)=\frac{\mathrm{p}(x)}{\mathrm{q}(x)}$.</p>
<p>It is further given that&nbsp;$\mathrm{p}'(3)=3$ and $\mathrm{q}'(3)=-1$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="q1-c" data-list-item-path="q1-c"><div class="list-item-paragraph">Explain why $\mathrm{r}^{-1}$ does not exist on the given domain.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="q1-d" data-list-item-path="q1-d"><div class="list-item-paragraph">Find the equation of the normal to $y=\mathrm{r}(x)$ at $x=3$.<span class="inline-marks">[3]</span></div></li></ol>

The functions $\mathrm{p}$ and $\mathrm{q}$ are continuous for $1\leq x\leq5$. The function $\mathrm{p}$ is strictly increasing, while $\mathrm{q}$ is strictly decreasing and $\mathrm{q}(x)>0$. Selected values are shown below.

The function $\mathrm{r}$ is defined by $\mathrm{r}(x)=\frac{\mathrm{p}(x)}{\mathrm{q}(x)}$.

It is further given that $\mathrm{p}'(3)=3$ and $\mathrm{q}'(3)=-1$.

Explain why $\mathrm{r}^{-1}$ does not exist on the given domain.

Find the equation of the normal to $y=\mathrm{r}(x)$ at $x=3$.

Composite Functions

One-to-One Functions and Inverse Existence

Tangents and Normals at a Point

Quotient Rule

Inverse Functions

Normal from tabulated function values

Functions

Techniques of Differentiation

Applications of Differentiation

<p>Mathematical correction: the stationary point $P$ is a maximum. The earlier wording said minimum; the equation and assessed parts are unchanged.</p>
<p>A curve $y=\dfrac{1+3\ln x}{x^2}$ has a maximum point at $P$.</p>
<ol class="abc-list"><li data-list-item-id="jvss-p1-q9-q-1-content-items-0" data-list-item-path="jvss-p1-q9-q-1-content-items-0"><div class="list-item-paragraph"><p>Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="jvss-p1-q9-q-1-content-items-1" data-list-item-path="jvss-p1-q9-q-1-content-items-1"><div class="list-item-paragraph"><p>Show that the $x$-coordinate of $P$ is $\mathrm{e}^{\frac{1}{6}}$.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="jvss-p1-q9-q-1-content-items-2" data-list-item-path="jvss-p1-q9-q-1-content-items-2"><div class="list-item-paragraph"><p>Find the equation of normal to the curve at the point $x=\mathrm{e}$. Leave your answer in terms of $\mathrm{e}$.</p><span class="inline-marks">[5]</span></div></li></ol>

Mathematical correction: the stationary point $P$ is a maximum. The earlier wording said minimum; the equation and assessed parts are unchanged.

A curve $y=\dfrac{1+3\ln x}{x^2}$ has a maximum point at $P$.

Derivatives of Natural Logarithmic Functions

Stationary Points and Their Nature

2025 JVSS PRELIMS P1 Q9

Differentiation

<p>A curve is given by the equation $y = \frac{3x + 10}{\sqrt{x + 1}}$.</p>
<ol class="abc-list"><li data-list-item-id="imported-item-block-1-item-0" data-list-item-path="imported-item-block-1-item-0"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence find the $x$-coordinate of the turning point of the curve.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-1-item-1" data-list-item-path="imported-item-block-1-item-1"><div class="list-item-paragraph">Use calculus to find the value of the second derivative at the turning point and state the nature of this turning point.<span class="inline-marks">[3]</span></div></li></ol>

A curve is given by the equation $y = \frac{3x + 10}{\sqrt{x + 1}}$.

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence find the $x$-coordinate of the turning point of the curve.

Use calculus to find the value of the second derivative at the turning point and state the nature of this turning point.

Second and Higher Derivatives

Chain Rule

N2025 P1 Q4

<ol class="i-list"><li data-list-item-id="amint-q12-question-1" data-list-item-path="amint-q12-question-1"><div class="list-item-paragraph">Show that&nbsp;$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\sin3x}{1-\cos3x}\right)=-\dfrac{3}{1-\cos3x}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="amint-q12-question-2" data-list-item-path="amint-q12-question-2"><div class="list-item-paragraph">Hence evaluate the exact value of $\displaystyle\int_{\pi/12}^{\pi/6}\dfrac{1}{1-\cos3x}\,\mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol>

Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\sin3x}{1-\cos3x}\right)=-\dfrac{3}{1-\cos3x}$.

Hence evaluate the exact value of $\displaystyle\int_{\pi/12}^{\pi/6}\dfrac{1}{1-\cos3x}\,\mathrm{d}x$.

Derivatives of Trigonometric Functions

Definite Integration as the Reverse of Differentiation

Sine quotient with a triple angle

Integration

<ol class="i-list"><li data-list-item-id="amint-q11-question-1" data-list-item-path="amint-q11-question-1"><div class="list-item-paragraph">Show that&nbsp;$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\sin2x}{1+\cos2x}\right)=\dfrac{2}{1+\cos2x}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="amint-q11-question-2" data-list-item-path="amint-q11-question-2"><div class="list-item-paragraph">Hence evaluate the exact value of $\displaystyle\int_0^{\pi/6}\dfrac{1}{1+\cos2x}\,\mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol>

Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\sin2x}{1+\cos2x}\right)=\dfrac{2}{1+\cos2x}$.

Hence evaluate the exact value of $\displaystyle\int_0^{\pi/6}\dfrac{1}{1+\cos2x}\,\mathrm{d}x$.

Sine quotient with a double angle

<ol class="i-list"><li data-list-item-id="amint-q09-question-1" data-list-item-path="amint-q09-question-1"><div class="list-item-paragraph">Show that&nbsp;$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{3x+2}{\sqrt{x+4}}\right)=\dfrac{3x+22}{2(x+4)^{3/2}}$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="amint-q09-question-2" data-list-item-path="amint-q09-question-2"><div class="list-item-paragraph">Hence evaluate $\displaystyle\int_{-3}^{5}\dfrac{3x+22}{2(x+4)^{3/2}}\,\mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol>

Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{3x+2}{\sqrt{x+4}}\right)=\dfrac{3x+22}{2(x+4)^{3/2}}$.

Hence evaluate $\displaystyle\int_{-3}^{5}\dfrac{3x+22}{2(x+4)^{3/2}}\,\mathrm{d}x$.

Linear numerator over a square root

<p>Given that&nbsp;$y = \frac{3\ln 2x}{x^2}$ for&nbsp;$x &gt; 0$.</p>
<ol class="abc-list"><li data-list-item-id="item-1781506047585-uk90uikz0"><div class="list-item-paragraph"> Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{x^3}(1 - 2\ln 2x)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1781506048125-wavvo5cvu"><div class="list-item-paragraph">Hence, find $\int \frac{\ln 2x}{x^3} \mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol>

Given that $y = \frac{3\ln 2x}{x^2}$ for $x > 0$.

Integration as an Inverse Process of Differentiation

2024 FMSS PRELIMS P2 Q3

<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\ln 2x}{3x^2} \right) = \frac{1}{3x^3} - \frac{2\ln 2x}{3x^3}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Integrate $\frac{\ln 2x}{x^3}$ with respect to $x$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Given that the curve $y = \mathrm{f}(x)$ passes through the point $ \left(\frac{1}{2}, 1 \right)$ and is such that&nbsp;$\mathrm{f}'(x) = \frac{\ln 2x}{x^3}$, find $\mathrm{f}(x)$.<span class="inline-marks">[3]</span></div></li></ol>

Finding the Equation of the Curve

2024 CGSS P1 Q12

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Given that&nbsp;$ \mathrm{f}(x) = \frac{2x + 3}{\sqrt{4x - 3}}$, find $ \mathrm{f}'(x)$.</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find the set of values of $x$ for&nbsp;$y = \ln\left(\frac{3 + 8x}{8x - 5}\right)$ to be a decreasing function.</div></li></ol>

Increasing and Decreasing Functions

2024 CCHY P1 Q4

<p>A curve $C$ has parametric equations</p>
<p style="text-align: center">\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for}\hspace{0.5em} \lambda &gt; 0.\]</p>
<p>The point $P$ is a variable point on $C$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">With reference to the origin $O$, $OP$ forms the diagonal of the rectangle $OQPR$, where vertices $Q$ and $R$ lie on the $x$- and $y$-axis respectively. Using differentiation, find the value of $\lambda$ which maximises the area of rectangle $OQPR$. You need to show that your answer gives a maximum.<span class="inline-marks">[5]</span></div></li></ol>
<p>When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence determine the rate of change of the $y$-coordinate of the point $P$ at this instant.<span class="inline-marks">[4]</span></div></li></ol>

\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for } \lambda > 0.\]

The point $P$ is a variable point on $C$.

When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.

Connected Rates of Change

Differentiation of Parametric Equations

Problems Involving Maximisation and Minimisation

2024 RI Promo Q4

<p>The curve $C$ has equation</p>
<p style="text-align: center">\[y = x - 2 + \frac{3k(x - 2)}{x^2 - 2k}, \quad \text{where}\hspace{0.5em} k &gt; 0, \ k \ne 2.\]</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Without the use of a calculator, find the coordinates of any points of intersection of $C$ with the axes.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Write down the equations of the asymptotes of $C$ in terms of $k$.<span class="inline-marks">[2]</span></div></li></ol>
<p>It is now given that&nbsp;$k = 1$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Show that the $x$-coordinates of the turning points of $C$ satisfy the equation \[x^4 - 7x^2 + 12x - 2 = 0.\] Hence find these $x$-coordinates.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Sketch $C$, labelling all essential features.<span class="inline-marks">[3]</span></div></li></ol>

\[y = x - 2 + \frac{3k(x - 2)}{x^2 - 2k}, \quad \text{where } k > 0, \ k \ne 2.\]

Sketching Rational Functions

Asymptotes and Discontinuities

2024 EJC Promo Q12 

Graphing Techniques

<ol class="abc-list" start="2"><li data-list-item-id="item-3b-list" data-list-item-path="item-3b-list"><ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-3b-list/item-i"><div class="list-item-content"><span class="list-item-text">The equation of a curve is $y = \frac{2x}{1 - x^2}$. Find an expression for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-3b-list/item-ii"><div class="list-item-paragraph">Using the result in <strong>(i)</strong> above, explain, with working, why the curve $y = \frac{2x}{1 - x^2}$ has no turning points.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

HSS 4Express AM MYE 2021

<ol class="num-list" start="2"><li><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Differentiate $\frac{5x^2 + 3}{10x^2 - 6}$ with respect to $x$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Differentiate $\left( \frac{5x^2 + 3}{10x^2 - 6} \right)^4$ with respect to $x$, leaving your answer in the form $\frac{kx(5x^2 + 3)^m}{(5x^2 - 3)^n}$ where&nbsp;$k$, $m$ and $n$ are constants.</span><span class="inline-marks">[2]</span></div></li></ol></li></ol>

XHS 2022 4Express AM MYE 

<p>A company buys $p\%$ of its electronic components from supplier $A$ and the remaining $(100 - p)\%$ from supplier $B$. The probability that a randomly chosen component supplied by $A$ is faulty is $0.05$. The probability that a randomly chosen component supplied by $B$ is faulty is $0.03$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$p = 25$, find the probability that a randomly chosen component is faulty.</span></div></li><li><div class="list-item-paragraph">For a general value of $p$, the probability that a randomly chosen component that is faulty was supplied by $A$ is denoted by $\mathrm{f}(p)$. Show that&nbsp;$\mathrm{f}(p) = \frac{0.05p}{0.02p + 3}$. Prove by differentiation that&nbsp;$\mathrm{f}$ is an increasing function for&nbsp;$0 \leq p \leq 100$, and explain what this statement means in the context of the question.</div></li></ol>

A company buys $p\%$ of its electronic components from supplier $A$ and the remaining $(100 - p)\%$ from supplier $B$. The probability that a randomly chosen component supplied by $A$ is faulty is $0.05$. The probability that a randomly chosen component supplied by $B$ is faulty is $0.03$.

Given that $p = 25$, find the probability that a randomly chosen component is faulty.

Bayes' Theorem

Law of Total Probability

N2009 P2 Q7

Probability

<p>The equation of a curve is $y=\frac{ \ln x}{x}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1773466434665-oon3h3sb1"><div class="list-item-paragraph">Find the coordinates of the stationary point of this curve. <br>Leave your answer in terms of e.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1773466436825-f0ta69f8n"><div class="list-item-paragraph">Determine the nature of the stationary point.<span class="inline-marks">[3]</span></div></li></ol>

The equation of a curve is $y=\frac{\ln x}{x}$.

Find the coordinates of the stationary point of this curve. Leave your answer in terms of $e$.

Determine the nature of the stationary point.

2025 MSHS PRELIMS P2 Q2

<ol class="abc-list"><li data-list-item-id="ahs-p2-q11-q-0-content-items-0" data-list-item-path="ahs-p2-q11-q-0-content-items-0"><div class="list-item-paragraph">Given the function $y=\frac{m-2x}{x^2-3}$, find $\frac{dy}{dx}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="ahs-p2-q11-q-0-content-items-1" data-list-item-path="ahs-p2-q11-q-0-content-items-1"><div class="list-item-paragraph">Hence, find the range of values of $m$ such that the function $y=\frac{m-2x}{x^2-3}$ is increasing on each of the intervals $( -\infty,-\sqrt3)$, $(-\sqrt3,\sqrt3)$ and $(\sqrt3,\infty)$ of its domain.<span class="inline-marks">[4]</span></div></li></ol>

Given the function $y=\frac{m-2x}{x^2-3}$, find $\frac{dy}{dx}$.

Hence, find the range of values of $m$ such that the function $y=\frac{m-2x}{x^2-3}$ is increasing on each of the intervals $( -\infty,-\sqrt3)$, $(-\sqrt3,\sqrt3)$ and $(\sqrt3,\infty)$ of its domain.

2025 AHS PRELIMS P2 Q11

<p>Differentiate with respect to $x$ of the following expressions leaving your answers in terms of $x$:</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">$\frac{\sec (\ln x)}{x}$</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">$\frac{e^{-2x}}{5^x}$</div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">$\ln \sqrt{e^{1+\sec x}}$</div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">$x^\pi$</div></li><li data-list-item-id="item-e" data-list-item-path="item-e"><div class="list-item-paragraph">$\pi^x$</div></li><li data-list-item-id="item-f" data-list-item-path="item-f"><div class="list-item-paragraph">$2^{\ln x}$</div></li><li data-list-item-id="item-g" data-list-item-path="item-g"><div class="list-item-paragraph">$5^{\cos x}$</div></li><li data-list-item-id="item-h" data-list-item-path="item-h"><div class="list-item-paragraph">$x^x$</div></li><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$(\ln x)^x$</div></li></ol>

Differentiate with respect to $x$ of the following expressions leaving your answers in terms of $x$:

Logarithmic Differentiation

Derivatives of Reciprocal Trigonometric Functions

Derivatives of Exponential Functions

Power Rule and Derivatives of Algebraic Functions

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">If&nbsp;$y = \frac{x + 2}{\sqrt{3x + 2}}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$.</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Use your result from <strong>(a)</strong> to find $\displaystyle\displaystyle\int_{2}^{4} \frac{3x + 6}{2 \sqrt{(3x + 2)^3}} \mathrm{d}x$.</div></li></ol>

If $y = \frac{x + 2}{\sqrt{3x + 2}}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Use your result from **(a)** to find $\displaystyle\displaystyle\int_{2}^{4} \frac{3x + 6}{2 \sqrt{(3x + 2)^3}} \mathrm{d}x$.

<p>The equation of a curve is $y = \frac{\sin x - 2}{\cos x}.$</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Obtain an expression for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">The point $P$ moves along the curve such that its $x$-coordinate is decreasing at a constant rate of $0.2$ unit/s. Find the rate of change of the $y$-coordinate $P$ when&nbsp;$x = \frac{\pi}{4}$.</div></li></ol>

The equation of a curve is $y = \frac{\sin x - 2}{\cos x}.$

Obtain an expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$.

The point $P$ moves along the curve such that its $x$-coordinate is decreasing at a constant rate of $0.2$ unit/s. Find the rate of change of the $y$-coordinate $P$ when $x = \frac{\pi}{4}$.

<p>The equation of a curve is $ y = \frac{2 \ln x}{x} . $</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Obtain an expression for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">The point $P$ moves along the curve such that its $x$-coordinate is increasing at a constant rate of $0.03$ unit/s. Find the rate of change of the $y$-coordinate of $P$ when&nbsp;$x = 1$.</div></li></ol>

The equation of a curve is $ y = \frac{2 \ln x}{x} . $

The point $P$ moves along the curve such that its $x$-coordinate is increasing at a constant rate of $0.03$ unit/s. Find the rate of change of the $y$-coordinate of $P$ when $x = 1$.

Math Question Bank