<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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My son was blown away by his first tutorial with Jamie. He loved Jamie's enthusiasm and said that everything was explained so well that concepts he's found difficult for a term and a half at school were quickly made clear. Also really liked the way that Jamie kept testing his understanding with questions. Fantastic, many thanks.

Tanya, Parent from Bucks

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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.

To answer this we will use implicit differentiation with respect to x. So start by differentiating each term. On the left hand side 0 differentiates to 0. On the right hand side 5x^{2} differentiates to 10x. By using the product rule and implicit differentiation 3xy differentiates to 3x dy/dx +3y. -y^{3} differentiates to -3y^{2} dy/dx by implicit differentiation. So the whole differentiated equation is 0=10x+3x dy/dx +3y - 3y^{2} dy/dx.

Then rearrange the equation so all terms containing dy/dx are on one-side of the equals sign and the other terms are on the other-side so 3y^{2} dy/dx -3x dy/dx = 10x+3y. Then take out a factor of dy/dx from the left hand side giving dy/dx(3y^{2}-3x)=10x+3y. Finally, divide each side by 3y^{2}-3x to get an equation in terms of dy/dx, dy/dx=(10x+3y)/(3y^{2}-3x). Then plug in the co-ordinates given above to obtain dy/dx=-7/9

Then rearrange the equation so all terms containing dy/dx are on one-side of the equals sign and the other terms are on the other-side so 3y

Answered by Holly W.

Studies Mathematics at Manchester

Since f(x) is a product of the two functions e^x and sin(x^2), we can use the product rule which states that if f(x)=g(x)h(x), then f'(x)=g'(x)h(x)+g(x)h'(x). Let g(x)=e^x and h(x)=sin(x^2). Since the differential of e^x is e^x, g'(x)h(x)=e^x sin(x^2), which is the first part of f'(x). For the second part of f'(x), g(x)h'(x), e^x is not differentiated, but we must use the chain rule to differentiate sin(x^2). The chain rule states that if h(x)=u(v(x)), then h'(x)=v'(x)u'(v(x)). Let u(x)=sin(x) and v(x)=x^2, differentiating x^2 using the power rule gives v'(x)=2x, and differentiating sin(x) gives u'(x)=cos(x), so u'(v(x))=cos(x^2), and h'(x)=v'(x)u'(v(x))=2xcos(x^2). This means that g(x)h'(x)=e^x 2xcos(x^2), so f'(x)=e^x sin(x^2)+e^x 2xcos(x^2).

Answered by Michael B.

Studies MPhys Mathematical Physics at Edinburgh

ln is the natural log. The thing to remember with differentiating natural log is the simple formula U'/U. The U is whatever is in the brackets. This means we differentiate X^2 and divide it by X^2. X^2 differentiated is 2X because of the chain rule. Now we have 2X/X^2. This simplifies to 2/X .

Answered by Edward T.

Studies Electrical and Electronic Engineering at Imperial College London

To differentiate y, we must used the product rule.The product rule is d/dx [f(x)g(x)] = f'(x)g(x) + g'(x)f(x)So here, we let f(x)= x^3 and g(x)= sin(x)Then, f'(x)= 3x^2 and g'(x) = cos(x)Then substituting these into the product rule formula, we get

dy/dx = (3x^2)sin(x) + cos(x)x^3

We can simplify the answer by factorising out x^2 :dy/dx= x^2[3sin(x) + xcos(x)]

dy/dx = (3x^2)sin(x) + cos(x)x^3

We can simplify the answer by factorising out x^2 :dy/dx= x^2[3sin(x) + xcos(x)]

Answered by Kajal C.

Studies Mathematics MSci at Bristol

6cos(2x) +sin(x).Using the double angle formula for cosine (or otherwise), cos(2x) = cos(x)cos(x) - sin(x)sin(x) .cos(2x) = cos^2(x) - sin^2(x) .Hence, 6cos(2x) +sin(x) = 6(cos^2(x) - sin^2(x)) + sin(x). Now use the trigonometric identity 1 = cos^2(x) + sin^2(x).6(cos^2(x) - sin^2(x)) + sin(x) = 6((1-sin^2(x)) - sin^2(x)) + sin(x) .6((1-sin^2(x)) - sin^2(x)) + sin(x) = 6 (1 - 2sin^2(x)) +sin(x) .Therefore, 6cos(2x) +sin(x) = 6 + sin(x) -12sin^2(x).**6cos(2x) +sin(x) = (4sin(x) − 3)*(3sin(x) + 2) **

Answered by Robbie M.

Studies Mechanical Engineering at Bristol

f’(x)= -6cos(3x)sin(3x)

Answered by Jeff M.

Studies Mathematics at Queen's, Belfast

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