Below, you will find features that will be included with a membership on L.C. Maths from the 2023/24 school year onwards.

Note that these are already live on A.M. Online, our Leaving Cert Applied Maths website. You can check out the free lessons on A.M. Online to see what the style of the lessons will be like here also!

Great question!

We started working full time creating that content across all three of our websites in the summer of 2017 with the hopes of finishing it by 2020.

We instead finished in January 2022.

However, yes, you read that correctly. We actually do have all of the content completed for this website as well!

However, there was still one important step that we were yet to complete on any of our three websites - quality checks. We were yet to quadruple check our lessons for grammar mistakes, quadruple check our solutions for maths errors etc.

We don't want our first students to be the ones finding those mistakes for us!

When we started those checks in January, we assumed that it would take approximately 2 months to perform those checks on each website, thereby allowing us to have everything completed by the summer.

However, even with some colleagues helping us, we found that this quality check process was taking longer than we expected. A lot longer in fact.

This is primarily due to the fact that our original plan was to make the content on each website as long as a typical schoolbook. Instead, the content on each website was almost ten times longer! 

We therefore decided to prioritise getting A.M. Online finished first for two reasons:

  • Many applied maths students study the subject from home and therefore a website like A.M. Online is of particular benefit to them.
  • The applied maths syllabus is changing for the 2023 Leaving Cert onwards. Therefore, there is a lack of online resources and support for all applied maths students for the new syllabus.

By prioritising A.M. Online, we completed these quality checks in August 2022, taking over half a year for just one website.

We are therefore spending the 2022/23 school year repeating that very same process for our other two websites with the goal of having them ready to go for the 2023/24 school year!

INTERACTIVE LESSONS

L.C. Maths is not simply a collection of teacher notes and recorded videos.

Instead, this one-of-a-kind, world-class website is composed of professional, interactive lessons that make learning Leaving Cert Maths an enjoyable experience for students.

Schoolbooks take a “fill in the gaps” approach, whereby a teacher is expected to expand on the core concepts of the book.

As we are not limited by what can fit into a schoolbook, our lessons instead describe every concept within Leaving Cert Maths with as much detail as possible.

Throughout each lesson, we also highlight three important aspects:

  • Key Points – the most important parts of the lesson.
  • Causes of Confusion – common mistakes that students make.
  • Exam Tips – tips to give you that extra edge on the exam!

Key Point

Quadratic expressions are expressions of the form

\begin{align}ax^2+bx+c\end{align}

where \(a\), \(b\) and \(c\) are constants.

Cause of Confusion

Not all quadratic expressions can be factored using the concept of factor pairs.

Exam Tip

If you’re ever unsure as to whether you factored a quadratic expression correctly or not, you can always double-check by expanding your answer.

ANIMATIONS

Animations make for a much more helpful and beneficial learning experience compared to the static images of a schoolbook.

We therefore provide rich, unique animations, together with thousands of illustrations, throughout the lessons in order to provide students with what we believe are the best learning resources available for Leaving Cert Maths.

Note: Animations, equations and indeed many other advanced features of this website may not render correctly on browser versions released way back in 2017 or earlier. In the unlikely event that you are still using such an outdated version, updating the browser will get everything working smoothly!

FULL SOLUTIONS

Schoolbooks typically only provide the final answers to practice questions in the back pages rather than the full details of how those answers are obtained.

Wouldn’t it again be awesome if your book instead provided those full details?

As you might have guessed, L.C. Maths has got that covered!

We provide detailed, written solutions to EVERY exercise question, and we have just as many questions to practice with as you’d find in a schoolbook!

Can you imagine having a detailed written solution for every single exercise question in your schoolbook?! Wow!

In fact, we go even further! We provide a total of four different methods for students to check their understanding of every question:

  • First, check to see if you got the right Answer. If you didn’t…
  • Read the Solution, i.e. what you should write down when answering the question. If you don’t understand the solution…
  • Read the Walkthrough, i.e. what a teacher would say if they were explaining the solution to you. If you don’t understand the walkthrough…
  • Ask Mr. Kenny for further guidance on Teacher Chat!

These “stepping stones” greatly reduce the likelihood of a student “giving up” if they don’t understand the solution to a question.

Note: The equations on this page may load slowly (or not at all) on certain mobile/tablet devices. Don’t worry! This page is very resource-heavy compared to a typical lesson page that you’ll find on this website. As long as everything loads correctly on e.g. this preview lesson page on A.M. Online, that device will work fine!

Practice Question

Consider the following relation:

\begin{align}y(x)=x^3\end{align}

(a) Represent this relation on a graph in the domain \(-1\leq x \leq 1\).

(b) Using part (a), graph the following relations in the same range:

(i) \(y(x) = -x^3\)

(ii) \(y(x) = \left(x+\dfrac{1}{2}\right)^3\)

(iii) \(y(x) = -\left(x+\dfrac{1}{2}\right)^3\)

(a)

yx0.510.5–1–0.5–10.51

(b)

(i)

yx0.510.5–1–0.5–10.51

(ii)

yx0.510.5–1–0.5–10.51

(iii)

yx0.510.5–1–0.5–10.51

(a) Factor pairs: \((-1,-1)\), \((-0.5,-0.125)\), \((0,0)\), \((0.5,0.125)\) and \((1,1)\).

yx0.510.5–1–0.5–10.51

(b)

(i)

yx0.510.5–1–0.5–10.51

(ii)

yx0.510.5–1–0.5–10.51

(iii)

yx0.510.5–1–0.5–10.51

(a) As we are told to restrict the domain to \(-1\leq x\leq 1\), we have chosen the following ordered pairs:

\begin{align}(-1,-1) && (-0.5,-0.125)\end{align}

\begin{align}(0,0)\end{align}

\begin{align}(0.5,0.125) && (1,1)\end{align}

Plotting these five points and drawing the resulting curve, we obtain the following graph.

yx0.510.5–1–0.5–10.51

(b) 

(i) This curve corresponds to reflecting the curve from part (a) through the \(x\) axis. 

yx0.510.5–1–0.5–10.51

(ii) This curve corresponds to shifting the curve from part (a) to the left \(0.5\) units.

yx0.510.5–1–0.5–10.51

(iii) This curve corresponds to shifting the curve from part (a) to the left \(0.5\) units

yx0.510.5–1–0.5–10.51

and then reflecting this new curve through the \(x\) axis.

yx0.510.5–1–0.5–10.51

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

PAST PAPER SOLUTIONS

Official past papers, and their marking schemes, are freely available online.

However, these marking schemes are often hard to follow as they do not fully explain how to go from one step to the next.

As such, we provide our own unique, detailed solutions to past examinations papers, both at Higher and Ordinary Level.

In fact, we provide the same four “stepping stones” as we provide for our practice questions, including solutions and walkthroughs!

We don’t just provide these solutions for past papers of the last few years. Instead, we provide our “stepping stones” for every single past paper question of the current syllabus.

For example, the question shown is Higher Level Paper 1 Question 1 from all the way back in 2015!

We organise these past paper questions both 1) by year so that students can practice with full papers and 2) by topic so that students can improve their understanding in specific areas.

Past Paper Question

Mary threw a ball onto level ground from a height of \(2 \mbox{ m}\). Each time the ball hit the ground it bounced back up to \(\frac{3}{4}\) of the height of the previous bounce, as shown.

Bounce2 m0Height

(a) Complete the table below to show the maximum height, in fraction form, reached by the ball on each of the first four bounces.

Bounce \(0\) \(1\) \(2\) \(3\) \(4\)

Height (m)

\(\dfrac{2}{1}\)

(b) Find, in metres, the total vertical distance (up and down) the ball had travelled when it hit the ground for the \(5\)th time. Give your answer in fraction form

(c) If the ball were to continue to bounce indefinitely, find, in metres, the total vertical distance it would travel.

(a)

Bounce \(0\) \(1\) \(2\) \(3\) \(4\)

Height (m)

\(\dfrac{2}{1}\)

\(\dfrac{3}{2}\)

\(\dfrac{9}{8}\)

\(\dfrac{27}{32}\)

\(\dfrac{81}{128}\)

(b) \(10\dfrac{13}{64}\mbox{ m}\)

(c) \(14\mbox{ m}\)

(a)

Bounce \(0\) \(1\) \(2\) \(3\) \(4\)

Height (m)

\(\dfrac{2}{1}\)

\(\dfrac{3}{2}\)

\(\dfrac{9}{8}\)

\(\dfrac{27}{32}\)

\(\dfrac{81}{128}\)

(b)

\begin{align}d &= \frac{2}{1} + 2\left(\frac{3}{2} + \frac{9}{8}+\frac{27}{32}+\frac{81}{128}\right)\\&= 2+2\left(\frac{192}{128}+\frac{144}{128}+\frac{108}{128}+\frac{81}{128}\right)\\&=2+2\left(\frac{525}{128}\right)\\&=2+\frac{1{,}050}{128}\\&=\frac{256}{128}+\frac{1{,}050}{128}\\&=\frac{1{,}306}{128}\\&=\frac{653}{64}\\&=10\frac{13}{64}\mbox{ m}\end{align}

(c)

\begin{align}d_\infty &= \frac{2}{1} + 2\left(\frac{3}{2} + \frac{9}{8}+\frac{27}{32}+\frac{81}{128}+…\right)\\&=2+2\left(\frac{\frac{3}{2}}{1-\frac{3}{4}}\right)\\&=2+2(6)\\&=14\mbox{ m}\end{align}

(a) With each successive bounce, the maximum height of the ball decreases by a factor of \(\dfrac{3}{4}\).

The initial height of the ball is \(2\mbox{ m}\), i.e. \(\dfrac{2}{1}\mbox{ m}\) in fraction form.

After the first bounce, the maximum height of the ball will then be three quarters of this height.

\begin{align}\frac{3}{4}\times \frac{2}{1} &= \frac{3\times 2}{4\times 1}\\&=\frac{6}{4}\mbox{ m}\\&=\frac{3}{2}\mbox{ m}\end{align}

Again, after the second bounce, the maximum height of the ball will be then three quarters of the previous maximum height.

\begin{align}\frac{3}{4}\times \frac{3}{2} &= \frac{3\times 3}{4\times 2}\\&=\frac{9}{8}\mbox{ m}\end{align}

Again, after the third bounce, the maximum height of the ball will then be three quarters of the previous maximum height.

\begin{align}\frac{3}{4}\times \frac{9}{8} &= \frac{3\times 9}{4\times 8}\\&=\frac{27}{32}\mbox{ m}\end{align}

Finally, after the fourth bounce, the maximum height of the ball will then be three quarters of the previous maximum height.

\begin{align}\frac{3}{4}\times \frac{27}{32} &= \frac{3\times 27}{4\times 32}\\&=\frac{81}{128}\mbox{ m}\end{align}

The table is therefore filled out as below.

Bounce \(0\) \(1\) \(2\) \(3\) \(4\)

Height (m)

\(\dfrac{2}{1}\)

\(\dfrac{3}{2}\)

\(\dfrac{9}{8}\)

\(\dfrac{27}{32}\)

\(\dfrac{81}{128}\)

(b) Before the first bounce, the ball falls a vertical distance of \(2\) metres.

From there, in between each bounce, the ball travels to its maximum height and then back down to the ground. Hence, between bounces, the vertical distance that the ball travels is twice the maximum height.

Therefore, with the exception of the first entry, we need to multiply each height in the above table by \(2\) in order to sum the vertical distances.

\begin{align}d &= \frac{2}{1} + 2\left(\frac{3}{2} + \frac{9}{8}+\frac{27}{32}+\frac{81}{128}\right)\\&= 2+2\left(\frac{192}{128}+\frac{144}{128}+\frac{108}{128}+\frac{81}{128}\right)\\&=2+2\left(\frac{525}{128}\right)\\&=2+\frac{1{,}050}{128}\\&=\frac{256}{128}+\frac{1{,}050}{128}\\&=\frac{1{,}306}{128}\\&=\frac{653}{64}\\&=10\frac{13}{64}\mbox{ m}\end{align}

(c) If the ball continues to bounce indefinitely, decreasing its maximum height by \(\dfrac{3}{4}\) each time, then the term in brackets from part (b) then becomes an infinite sum.

\begin{align}d_\infty &= \frac{2}{1} + 2\left(\frac{3}{2} + \frac{9}{8}+\frac{27}{32}+\frac{81}{128}+…\right)\end{align}

From part (a), we know that there is a constant ratio of \(\dfrac{3}{4}\) between successive terms of the expression within the brackets. This expression is therefore an infinite geometric series of the form

\begin{align}S_\infty = \frac{a}{1-r}\end{align}

In this case, the first term within the bracketed expression is \(a=\dfrac{3}{2}\) and again, according to part (a), the common ratio between successive terms in the bracketed expression is \(r=\dfrac{3}{4}\).

\begin{align}S_\infty &= \frac{\frac{3}{2}}{1-\frac{3}{4}}\\&=\frac{\frac{3}{2}}{\frac{1}{4}}\\&=\frac{3}{2}\times \frac{4}{1}\\&=\frac{12}{2}\\&=6\end{align}

Therefore, we can simply replace the bracketed expression with \(6\)!

\begin{align}d_\infty &= \frac{2}{1} + 2\left(\frac{3}{2} + \frac{9}{8}+\frac{27}{32}+\frac{81}{128}+…\right)\\&=2+2(6)\\&=14\mbox{ m}\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

Sample Paper Question

The equations of two lines are shown below:

\begin{align}\mathbf{A}:4x-3y+15=0\end{align}

\begin{align}\mathbf{B}:ky=2x-4\end{align}

where \(k\in \mathbb{R}\).

(a) Assuming that the lines are parallel to each other, find:

(i) the value of \(k\).

(ii) the perpendicular distance between the lines.

(b) Assuming that the lines are perpendicular to each other, find:

(i) the value of \(k\).

(ii) the intersection point.

(c) If \(k=1\), find the acute angle between both lines, to the nearest degree.

(a)

(i) \(k=\dfrac{3}{2}\)

(ii) \(\dfrac{23}{5}\)

(b)

(i) \(-\dfrac{8}{3}\)

(ii) \(\left(-\dfrac{42}{25},\dfrac{181}{25}\right)\)

(c) \(22^{\circ}\)

\begin{align}\mathbf{A}:y=\frac{4}{3}x+5\end{align}

\begin{align}\mathbf{B}:y=\frac{2}{k}x-\frac{4}{k}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m_A = \frac{4}{3} && m_B=\frac{2}{k}\end{align}

(a)

(i)

\begin{align}\frac{4}{3}=\frac{2}{k}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4k=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=\frac{3}{2}\end{align}

\[\,\]

(ii) Line \(B\):

\begin{align}y(0)&=-4(0)-\frac{8}{3}\\&=-\frac{8}{3}\end{align}

\(\rightarrow \left(0,-\dfrac{8}{3}\right)\) is a point on line \(B\). Therefore:

\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{|(4)(0)+(-3)\left(-\frac{8}{3}\right)+15|}{\sqrt{4^2+(-3)^2}}\\&=\frac{|0+8+15|}{\sqrt{25}}\\&=\frac{23}{5}\end{align}

(b)

(i)

\begin{align}m_1\times m_2 = -1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4}{3}\times\frac{2}{k}=-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{8}{3k}=-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=-\frac{8}{3}\end{align}

\[\,\]

(ii)

\begin{align}\mathbf{A}:y=\frac{4}{3}x+5\end{align}

\begin{align}\mathbf{B}:y=-\frac{3}{4}x+\frac{3}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=\left[\frac{4}{3}x-\left(-\frac{3}{4}x\right)\right]+\left(5-\frac{3}{2}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=\frac{25}{12}x+\frac{7}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=-\frac{7}{2}\times \frac{12}{25}\\&=-\frac{84}{50}\\&=-\frac{42}{25}\end{align}

To find \(y\), we will use:

\begin{align}y&=\frac{4}{3}x+5\\&=\frac{4}{3}\left(\frac{42}{25}\right)+5\\&=\frac{168}{75}+5\\&=\frac{543}{75}\\&=\frac{181}{25}\end{align}

Therefore, the intersection point is

\begin{align}\left(-\frac{42}{25},\frac{181}{25}\right)\end{align}

(c)

\begin{align}\tan \theta &=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\\&=\left|\frac{\frac{4}{3}-2}{1-\frac{4}{3}\times 2}\right|\\&=\left|\frac{-\frac{2}{3}}{-\frac{5}{3}}\right|\\&=\frac{2}{5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta &= \tan^{-1}\left(\frac{2}{5}\right)\\&\approx 22^{\circ}\end{align}

First, let us rearrange both equation so that they are in the form \(y=mx+c\).

\begin{align}\mathbf{A}:y=\frac{4}{3}x+5\tag{1}\end{align}

\begin{align}\mathbf{B}:y=\frac{2}{k}x-\frac{4}{k}\tag{2}\end{align}

(a)

(i) If the lines are parallel to each other, they will have the same slope.

According to equations \((1)\) and \((2)\), the slope of line \(A\) is \(\dfrac{4}{3}\) and the slope of line \(B\) is \(\dfrac{2}{k}\).

Setting these equal to each other, we obtain:

\begin{align}\frac{4}{3}=\frac{2}{k}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4k=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=\frac{3}{2}\end{align}

\[\,\]

(ii) The equations of both lines are therefore now as follows:

\begin{align}\mathbf{A}:y=\frac{4}{3}x+5\end{align}

\begin{align}\mathbf{B}:y=\frac{4}{3}x-\frac{8}{3}\end{align}

To find the perpendicular distance between both lines, we need to first find any point \((x_1,y_1)\) on either line.

For example, for line \(B\):

\begin{align}y(0)&=-4(0)-\frac{8}{3}\\&=-\frac{8}{3}\end{align}

and therefore \(\left(0,-\dfrac{8}{3}\right)\) is a point on line \(B\).

The perpendicular distance between \(\left(0,-\dfrac{8}{3}\right)\) and line \(A\) can now be found.

As line \(A\) is already in form \(ax+by+c=0\) in the question, we can immediately use the usual perpendicular distance formula.

\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{|(4)(0)+(-3)\left(-\frac{8}{3}\right)+15|}{\sqrt{4^2+(-3)^2}}\\&=\frac{|0+8+15|}{\sqrt{25}}\\&=\frac{23}{5}\end{align}

(b)

(i) If the lines are perpendicular to each other, their slopes satisfy the following:

\begin{align}m_1\times m_2 = -1\end{align}

According to equations \((1)\) and \((2)\), the slope of line \(A\) is \(\dfrac{4}{3}\) and the slope of line \(B\) is \(\dfrac{2}{k}\).

Inserting these into the above relation, we obtain

\begin{align}\frac{4}{3}\times\frac{2}{k}=-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{8}{3k}=-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=-\frac{8}{3}\end{align}

\[\,\]

(ii) The equations of both lines are therefore now as follows:

\begin{align}\mathbf{A}:y=\frac{4}{3}x+5\end{align}

\begin{align}\mathbf{B}:y=-\frac{3}{4}x+\frac{3}{2}\end{align}

To find the intersection point, we need to solve these two equations simultaneously.

We can easily eliminate \(y\) by subtracting the second equation from the first.

\begin{align}0=\left[\frac{4}{3}x-\left(-\frac{3}{4}x\right)\right]+\left(5-\frac{3}{2}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=\frac{25}{12}x+\frac{7}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=-\frac{7}{2}\times \frac{12}{25}\\&=-\frac{84}{50}\\&=-\frac{42}{25}\end{align}

Now that we know \(x\), we can find \(y\) using any equation containing \(y\) from part (b), including

\begin{align}y&=\frac{4}{3}x+5\\&=\frac{4}{3}\left(\frac{42}{25}\right)+5\\&=\frac{168}{75}+5\\&=\frac{543}{75}\\&=\frac{181}{25}\end{align}

Therefore, the intersection point is

\begin{align}\left(-\frac{42}{25},\frac{181}{25}\right)\end{align}

(c) The acute angle \(\theta\) between two intersecting lines of slope \(m_1\) and \(m_2\) is determined by

\begin{align}\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|\end{align}

According to equations \((1)\) and \((2)\), the slope of line \(A\) is \(\dfrac{4}{3}\) and the slope of line \(B\) is \(\dfrac{2}{1}=2\).

Inserting these into the above equation, we obtain

\begin{align}\tan \theta &=\left|\frac{\frac{4}{3}-2}{1-\frac{4}{3}\times 2}\right|\\&=\left|\frac{-\frac{2}{3}}{-\frac{5}{3}}\right|\\&=\frac{2}{5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta &= \tan^{-1}\left(\frac{2}{5}\right)\\&\approx 22^{\circ}\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

SAMPLE PAPERS

Just in case past paper solutions aren’t enough, we also provide an additional TEN sample examination papers so that both Higher and Ordinary Level students don’t have to worry about not having enough exam questions to practice with!

These papers can only be found on L.C. Maths and, as you may have guessed by now, also come with the same four “stepping stones” for every question!

As with past papers, these sample paper questions are also organised both by year and by topic.

KNOWLEDGE CHECKS

Before jumping directly into answering full exercise questions, we first test each student’s basic understanding of the lesson content with quick-fire questions in the form of Knowledge Checks.

These questions give each student immediate feedback as to whether a lesson “clicked” or not, thereby indicating whether they are ready to answer the corresponding exercise set or whether they need to do some quick revision first.

/10

\[\,\]Knowledge Check 38

\[\,\]A quick check on your understanding of:

Derivatives by Rule

The Product Rule

The Quotient Rule

The Chain Rule

\[\,\]

The slope of a line has a constant value.

1 / 10

The derivative of a linear function is a constant.

The derivative of a linear function is equal to the constant slope of the corresponding line.

2 / 10

The constant value of \(\dfrac{dy}{dx}\) is largest for which of the following lines?

\[\,\]

\[\,\]

If \(\dfrac{dy}{dx}\) has a constant value, the function \(y(x)\) is represented by a line.

3 / 10

If \(\dfrac{dy}{dx} = 0\), then \(y(x)\) is represented by a vertical line on the \(x\)-\(y\) plane.

This is a function of the form \(ax^n\).

4 / 10

If \(y(x) = 3x^5\), then what is \(\dfrac{dy}{dx}\)?

This is a function of the form \(ax^n\).

5 / 10

If \(y(x) = \dfrac{x^3}{3}\), then what is \(\dfrac{dy}{dx}\)?

This function is of the form \(ax^n\).

6 / 10

To calculate the derivative of \(13x^4\), we need to use the product rule.

This is a ratio of two functions.

7 / 10

To calculate the derivative of \(\dfrac{\sin x}{x}\), we need to use the quotient rule.

This is a product of two functions.

8 / 10

To calculate the derivative of \(e^x\sin x\), we need to use the product rule.

This is an example of a function nested within another function.

9 / 10

To calculate the derivative of \(\cos(\sin x)\), we need to use the product rule.

This is an example of a function nested within another function.

10 / 10

To calculate the derivative of \(x^{\sin x}\), we need to use the chain rule.

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TEACHER CHAT

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ONLINE COURSE:

Interactive Lessons

Practice Questions

Animations

Sample Papers

Walkthrough Solutions

and more!

Next year, students can get access to all of the above until June 30th 2024 for the same regular membership price of €49.

CHEAP AND CHEERFUL

Similar websites typically charge hundreds of euro per year for an online course that contains significantly less features. These are businesses whose primary goal is to make a profit.

L.C. Maths, on the other hand, was created by a teacher whose sole ambition is to give all students access to the best leaving cert maths resources, all in one place. 

As such, even though our online course is more feature-rich than any other online course that you can find, our weekly grinds and our separate Teacher Chat feature will still also be included with a membership!

For the same price as a typical grind or two, we will therefore provide everything needed to ace the exam for an entire school year, all on one website.

Alternative L.C. Maths

Ask a teacher for help only during class.

Ask a teacher for help throughout the day, every day.

Only answers at the back of the schoolbook.

Full, detailed solutions to every question in addition to detailed walkthroughs.

Learn using dull, static images.

Learn using vibrant animations.

Attempt to fill the gaps left out from marking schemes.

Solutions and walkthroughs included both for past papers and sample papers.