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PSLE Foundation Mathematics

<p>The diagram shows part of the graph $y = x^{\cos 2x}$, for&nbsp;$0 \le x \le 5$, which represents the path of a roller coaster. The horizontal distance travelled by the roller coaster is denoted by $x$ units and its vertical distance travelled is denoted by $y$ units.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1783573222728-q8fnjduy6xq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODM1NzMyMjI3MjgtcThmbmpkdXk2eHEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mjc4OTI2MSwiZXhwIjoxNzkyNzg5NTYxfQ.aHO0dIhEJuDztw5Xgtp0rjAMFp7chOJuP4Y6U1HKsu4" style="max-width: min(280px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = x^{\cos 2x} \left[ \frac{\cos 2x}{x} - (2 \sin 2x) \ln x \right]$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">At the point on the graph where&nbsp;$x = \pi$, find the rate at which the roller coaster is moving vertically when it is moving horizontally at a rate of $8$ units per hour.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Find the acute angle that the tangent to the graph where&nbsp;$x = \pi$ makes with the horizontal.<span class="inline-marks">[1]</span></div></li></ol>

The diagram shows part of the graph $y = x^{\cos 2x}$, for $0 \le x \le 5$, which represents the path of a roller coaster. The horizontal distance travelled by the roller coaster is denoted by $x$ units and its vertical distance travelled is denoted by $y$ units.

Derivatives of Trigonometric Functions

Logarithmic Differentiation

Derivative as the Gradient of a Curve

Connected Rates of Change

2024 ACJC P2 Q1

Techniques of Differentiation

Applications of Differentiation

<p>It is given that&nbsp;$y = x^{xy}$, where&nbsp;$x &gt; 0$, $y &gt; 0$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2 (1 + \ln x)}{1 - \ln y}$.<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">Hence find the coordinates of the point on the curve $y = x^{xy}$ whose tangent is parallel to the $y$-axis.<span class="inline-marks">[2]</span></div></li></ol>

It is given that $y = x^{xy}$, where $x > 0$, $y > 0$.

Implicit Differentiation

Tangents and Normals Parallel to the Axes

Tangents and Normals to Implicit Curves

2025 ACJC P1 Q3

<p>It is given that&nbsp;$y = \mathrm{e}^{\tan x}$, for&nbsp;$-\frac{\pi}{2} &lt; x &lt; \frac{\pi}{2}$, and $y_r = \frac{\mathrm{d}^r y}{\mathrm{d}x^r}$, for&nbsp;$r = 1, 2, 3$.</p>
<ol class="abc-list"><li data-list-item-id="part-a" data-list-item-path="part-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{y_1}{y} = 1 + \tan^2 x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="part-b" data-list-item-path="part-b"><div class="list-item-paragraph">By expressing $\ln y_1$ in terms of $\tan x$, show that&nbsp;$\frac{y_2}{y_1} = (1 + \tan x)^2$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="part-c" data-list-item-path="part-c"><div class="list-item-paragraph">Find $\frac{y_3}{y_2}$ in terms of $\tan x$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="part-d" data-list-item-path="part-d"><div class="list-item-paragraph">Hence obtain the Maclaurin series for&nbsp;$\mathrm{e}^{\tan x}$ in ascending powers of $x$ up to and including the term in $x^3$.<span class="inline-marks">[2]</span></div></li></ol>

It is given that $y = \mathrm{e}^{\tan x}$, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$, and $y_r = \frac{\mathrm{d}^r y}{\mathrm{d}x^r}$, for $r = 1, 2, 3$.

Maclaurin Series from Differential Equations

Higher Derivatives by Implicit Differentiation

Derivatives of Exponential Functions

Chain Rule

Maclaurin Series and Power Series

<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 $x$:

Derivatives of Natural Logarithmic Functions

Quotient Rule

Derivatives of Reciprocal Trigonometric Functions

Power Rule and Derivatives of Algebraic Functions

<p>Given the following relations between $x$ and $y$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">${{y}^{x}}=y+x$ </div></li><li><div class="list-item-paragraph">$y={{x}^{\text{sec}\hspace{0.5em}x}}$ </div></li><li><div class="list-item-paragraph">$y={{x}^{{{x}^{2}}}}$ </div></li></ol>

Given the following relations between $x$ and $y$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.

<p>It is given that&nbsp;$y = x^{xy}$, where&nbsp;$x &gt; 0$, $y &gt; 0$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2 (1 + \ln x)}{1 - \ln y}$<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">Hence find the coordinates of the point on the curve $y = x^{xy}$ whose tangent is parallel to the $y$-axis.<span class="inline-marks">[2]</span></div></li></ol>

Show that \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2 (1 + \ln x)}{1 - \ln y}\]

Hence find the coordinates of the point on the curve $y = x^{xy}$ whose tangent is parallel to the $y$-axis.

<p>It is given that $y={{x}^{\sin x}}$ where $x>0$.</p>
<p>Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$.</p>

It is given that $y={{x}^{\sin x}}$ where $x>0$.

Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$.

Worked Example 2

<p>Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the following functions in terms of $x$:</p>
<ol class="i-list"><li><div class="list-item-paragraph">$y = x^{\sin x}$</div></li><li><div class="list-item-paragraph">$y = x^{\tan x}$</div></li><li><div class="list-item-paragraph">$y = (\cos x)^{\ln x}$</div></li></ol>

Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the following functions in terms of $x$:

<p>Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the following equations in terms of $x$ and $y$:</p>
<ol class="i-list"><li><div class="list-item-paragraph">$x^{3} + y^{2} + xy = \sin y$</div></li><li><div class="list-item-paragraph">$y^{3} - 2xy = \sin y$</div></li><li><div class="list-item-paragraph">$y = x^{x}$</div></li><li><div class="list-item-paragraph">$x^{2} + 4y^{2} = 3yx$</div></li><li><div class="list-item-paragraph">$5x^{3} - 3y^{3} = 3yx$</div></li><li><div class="list-item-paragraph">$\tan y + x^{2} = x\ln y$</div></li><li><div class="list-item-paragraph">$\cos^{-1}(e^{y}) = x + 2y$</div></li></ol>

Find the $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the following equations in terms of $x$ and $y$:

Derivatives of Inverse Trigonometric Functions

Implicit differentiation: additional notes

For a constant base, $\frac{\mathrm d}{\mathrm dx}(a^x)=a^x\ln a$; for a constant exponent, $\frac{\mathrm d}{\mathrm dx}(x^n)=nx^{n-1}$. When both the base and exponent vary, logarithmic differentiation is useful.

The related logarithmic rules are $\frac{\mathrm d}{\mathrm dx}(\ln x)=\frac1x$, $\frac{\mathrm d}{\mathrm dx}(\log_a x)=\frac1{x\ln a}$, and $\log_a b=\frac{\ln b}{\ln a}$.

For the example $y=x^x$, take natural logarithms to obtain $\ln y=x\ln x$.

Differentiating implicitly gives $\frac1y\frac{\mathrm dy}{\mathrm dx}=x\left(\frac1x\right)+\ln x=1+\ln x$.

Thus $\frac{\mathrm dy}{\mathrm dx}=y(1+\ln x)=x^x(1+\ln x)$.

Implicit Differentiation: Additional Notes

<ol class="abc-list"><li><div class="list-item-paragraph">Given that $y={{x}^{\cos x}}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ only.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">A curve has equation ${{x}^{2}}y-x{{y}^{2}}=6$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the exact coordinates of the point on the curve where $\frac{\mathrm{d}y}{\mathrm{d}x}=1$.</div><p style="text-align: right;">[5]</p></li></ol></li></ol>

2024 YIJC Promo Q4

<p>A curve has equation</p>
<p style="text-align: center">${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that ${y}$ is a solution of the following differential equation.</div><div class="list-item-paragraph" style="text-align: center">${{{x}^{3}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-x\left( 1-\ln x \right)\frac{\mathrm{d}y}{\mathrm{d}x}+\left( 3-2\ln x \right)y=0}$.</div></li><li><div class="list-item-paragraph">Show further that</div><div class="list-item-paragraph" style="text-align: center">${{{\left. \frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}} \right|}_{x=\mathrm{e}}}=-{{\mathrm{e}}^{{{\mathrm{e}}^{-1}}-3}}}$.</div></li></ol>

${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$.

Second-Order Linear Differential Equations

Verifying Solutions of Differential Equations

Techniques and Applications Of Differentiation Q1

Differential Equations

<ol class="i-list"><li><div class="list-item-paragraph">If $y=\ln \left( \frac{{{\mathrm{e}}^{\sqrt{x}}}}{{{\cos }^{3}}x} \right)$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$, where $0 < x < \frac{\pi }{2}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Given that ${{y}^{\frac{1}{x}}}={{x}^{\ln x}}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.</div><p style="text-align: right;">[3]</p></li></ol>

2022 ACJC Promo Q1

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Given that&nbsp;$y={{\left( 2x+1 \right)}^{x}}$ for&nbsp;$x&gt;0$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$.<span class="inline-marks">[3]</span></div></li></ol>

2012 Promo CJC Q11

<p>Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left[ {{\left( \sqrt{x} \right)}^{x}} \right]=\frac{1}{2}{{\left( \sqrt{x} \right)}^{x}}\left( 1+\ln x \right)$.</p><p style="text-align: right;">[4]</p>
<p>Hence by using the substitution $u=\sqrt{x}$, evaluate $\displaystyle\int_{1}^{2}{{{\left( u \right)}^{{{u}^{2}}+1}}}\left( 1+2\ln u \right)\ \mathrm{d}u$ exactly.</p><p style="text-align: right;">[4]</p>

Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left[ {{\left( \sqrt{x} \right)}^{x}} \right]=\frac{1}{2}{{\left( \sqrt{x} \right)}^{x}}\left( 1+\ln x \right)$.

Hence by using the substitution $u=\sqrt{x}$, evaluate $\displaystyle\displaystyle \displaystyle\int_{1}^{2}{{{\left( u \right)}^{{{u}^{2}}+1}}}\left( 1+2\ln u \right)\ \mathrm{d}u$ exactly.

Integration as an Inverse Process of Differentiation

Integration by Substitution

2018 DHS J2 MYCT P1 Q5

Applications of Integration

Techniques of Integration

<p>Given that curve $C$ is defined by ${{y}^{x}}=x$ for $x>0$,</p>
<ol class="i-list"><li><div class="list-item-paragraph">Verify that $C$ has a stationary point at $x=\mathrm{e}$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Show that $x\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+\frac{\mathrm{d}y}{\mathrm{d}x}\left( 2+\ln y-\frac{1}{x} \right)=-\frac{y}{{{x}^{2}}}$ and deduce the nature of the stationary point found in (i).</div><p style="text-align: right;">[5]</p></li></ol>

Given that curve $C$ is defined by ${{y}^{x}}=x$ for $x>0$,

Stationary Points and Their Nature

2022 TJC J1 MYE Q7

<p>Given that a curve has the equation ${{x}^{{{\tan }^{-1}}x}}={{y}^{x+1}}$ where $x>0$ and $y>0$, using a non-calculator method, find the exact gradient of the curve at the point where $x=1$.</p><p style="text-align: right;">[5]</p>

Given that a curve has the equation ${{x}^{{{\tan }^{-1}}x}}={{y}^{x+1}}$ where $x>0$ and $y>0$, using a non-calculator method, find the exact gradient of the curve at the point where $x=1$.

2025 RI Techniques of Differentiation Assignment 6A Q3

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left( {{3}^{2x}} \right)={{9}^{x}}\ln 9$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, differentiate ${{\tan }^{-1}}\left( {{3}^{2x}} \right)$, $x>0$ with respect to $x$.</span></div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">Given that ${{y}^{2y}}={{e}^{3x+2e}}$, where $y>{{e}^{-1}}$, show that $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=-\frac{2}{3}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}$.</span></div></li></ol>

Given that ${{y}^{2y}}={{e}^{3x+2e}}$, where $y>{{e}^{-1}}$, show that $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=-\frac{2}{3}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}$.

Second and Higher Derivatives

2021 SAJC Promo Q5

<p>Differentiate the following with respect to $x$, leaving your answers in term of $x$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">${{e}^{ex}}\cot \left( \frac{x}{2} \right)$</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">${{\left( 2x \right)}^{1/x}}$</span></div><p style="text-align: right;">[4]</p></li></ol>

Differentiate the following with respect to $x$, leaving your answers in term of $x$.

${{e}^{ex}}\cot \left( \frac{x}{2} \right)$

Product Rule

2010 ACJC Promo Q5

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given that $y={{\tan }^{-1}}\sqrt{x}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Given that $\sqrt[x]{y}=\sqrt[y]{x}$, where $x>0$, $y>0$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ .</span></div><p style="text-align: right;">[4]</p></li></ol>

Given that $y={{\tan }^{-1}}\sqrt{x}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Given that $\sqrt[x]{y}=\sqrt[y]{x}$, where $x>0$, $y>0$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ .

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