<p>If you're taking A-Level Maths - well done to you. It's one of the trickier A-levels but also one of the most well respected, and it will set you up for all sorts of career paths. That said, it can be a surprising jump up from the Maths you've done before. And because the course progresses so quickly, it's easy to feel like you're getting left behind.</p>
<p><strong>With one-to-one online support from a Maths tutor, you can spend time focusing on the topics you want to work on.</strong> Whether it's differentiation that's got you in a pickle, or logarithms that are giving you trouble, our tutors are on hand to help. So by the time it comes round to your exams, you'll feel confident, well prepared, and able to achieve the results you want.</p>

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Fantastic first session. My son came away very positive and thought it helped alot. Alexander was very dedicated to finding the best way to help my son who has been struggling in a level maths. thank you so much for a brilliant experience, great service, great value. Will be booking another session.

Carolyn, Parent from Wiltshire

Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your Online Lessons are recorded. Make the most out of your live session, then play it back after

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after Online Lessons. In my Online Lessons, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the Online Lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

In order to solve this question we simply must substitute x=1 into f(x). If we carry out this substitution we see that
f(1) = 2(1^3) + (1^3) -5(1) + c = 2(1) + (1) -5(1) + c = 2 + 1 - 5 + c = 3 - 5 + c = c - 2.
We also know from the question provided that f(1) = 0. We can therefore match this condition with the substitution that we have just made which allows us to make the following statement.
f(1) = c - 2 = 0.
We can then rearrange this equation in order to get an expression for c by adding 2 to both sides, therefore c = 2.

Answered by Michael L.

Studies MSc Mathematics for Real-World Systems at Warwick

You can not integrate tan^{2}x but you can integrate sec^{2}x
Since *sec*^{2}x = 1 + tan^{2}x Then *tan*^{2}x = sec^{2}x-1
so the intragral of tan^{2}x dx = the integral of (sec^{2}x-1) dx = intrgral of sec^{2}x dx + integral of 1 dx
= tanx-x +C

Answered by Nandini P.

Studies Chemical Engineering at Edinburgh

find dy/dx, algebraically manipulate the expression to get dy/dx in terms of x and y and then subsitute in the given point.

-9=x^2+6x
0=x^2+6x+9
0=(x+3)(x+3)
when y=-9 x=3
dy/dx=2x+6
dy/dx=2(3)+6=12

Answered by Hassan M.

Studies MEDICINE at Imperial College London

When integrating, you need to add one to the power and divide the term by the power.
We will consider each term individually,
2x^{4} will become (2x^{4+1})/(4+1) = (2x^{5})/5,
-4x^{-0.5} will become (-4x^{-0.5+1})/(-0.5+1) = (-4x^{0.5})/(0.5) = -8x^{0.5} and
3 = 3x^{0} will become (3x^{0+1})/(0+1) = 3x.
Therefore,
∫ ( 2x^4 - 4x^(-0.5) + 3 ) dx = (2x^{5})/5 -8x^{0.5} + 3x + C,
where C is a constant of integration.
Since integration and differentiation are the inverse of each other, the C appears because there could have been a number which became zero when the formula was differentiated. Therefore, we must include a constant C when integrating.
You can check your answer because differentiating the answer will give you the formula within the integral.

A quick analysis here is based on the fact that y=(x^{2}). A big change is worked out between two points. The gradient between x=1 and x=2 is equal to 3. BUT we know the gradient is constantly changing so this is an average over a large change. This is not sufficient to model changes over the whole curve.

What we need to do is approximate in a space where the curve matches our quick "gradient = change in y over change in x" model as closely as possible. This is easy. If we zoom in on a curve enough, it will begin to look like a straight line. Don't believe me? The earth is curved if seen from space but if you zoom closely enough it appears flat.

so instead we`re going to look at the tiniest change in x, from x to (x+h), where h is tiny.

Then if:

y = x^{2}, then

gradient = ((x+h)^{2}-x^{2}) / ((x+h)-x)

= (x^{2} +2xh +h^{2} -x^2) / h

= (2xh+h^{2)} / h

= h(2x+h) / h

= (2x+h)

Now all we do is reduce the size of h until it reaches 0. So over no change in x value whatsoever. If y=x^{2} then the gradient dy/dx=2x.

We can explore this with other examples!

What we need to do is approximate in a space where the curve matches our quick "gradient = change in y over change in x" model as closely as possible. This is easy. If we zoom in on a curve enough, it will begin to look like a straight line. Don't believe me? The earth is curved if seen from space but if you zoom closely enough it appears flat.

so instead we`re going to look at the tiniest change in x, from x to (x+h), where h is tiny.

Then if:

y = x

gradient = ((x+h)

= (x

= (2xh+h

= h(2x+h) / h

= (2x+h)

Now all we do is reduce the size of h until it reaches 0. So over no change in x value whatsoever. If y=x

We can explore this with other examples!

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Time: | 2018-01-17T10:26:32Z |