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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 lessons. In my tutorials, 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 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.

Because we have a product of two functions of x, our first instinct is to apply integration by parts. Let u = e^(2x) and v' = cos(x). We then integrate v' to find v = sin(x) and differentiate u to find u' = 2e^(2x). Applying the by parts rule I = uv - (the integral of)(vu') we get I = e^(2x)*sin(x) - 2*(the integral of)(e^(2x)*sin(x)). The integral on the RHS is similar to the one we started with, so apply integration by parts again, this time with u = e^(2x), v' = sin(x), u' = 2e^(2x), v = -cos(x).
This gives us I = e^(2x)*sin(x) + 2e^(2x) - 4*(the integral of)(e^(2x)*cos(x)). The integral on the RHS is what we started with, so we substitute I in for it, getting I = e^(2x)*sin(x) + 2e^(2x) - 4I. Rearranging and solving for I gives us I = e^(2x)*(1/5)*(sin(x) + 2cos(x)).

(i): 15x^(2)+7 --- in order to arrive at this answer, we can divide the equation into 3 separate parts: 5x^(3) and 7x and 3. For the first part, you would multiply the first number, 5 by the power, in this case, 3, leaving us with 15. Then, you have to decrease the power by 2, leaving us with 15x^(2).
For the second part, the power is actually 1, so 7x^(1). The same process is used, multiply 7 by 1, leaving us with 7. Decrease the power by 1, leaving us with 0. Anything multiplied to the power of 0, is 1. 7x1=7.

Answered by Angela P.

Studies Law at Birmingham

3

Well, if you have n terms, we can prove that the sum of them is 1/(1-r) - r^n (1/(1-r)) - now as n gets bigger and bigger, the first bit doesn't change but the second bit gets smaller and smaller whenever |r|<1 - that's how the formula for infinite sums comes about algebraically. To give an intuitive answer, imagine you're on one side of a room and you want to walk across to the other side but you're only allowed to walk half-way between where you are and where you want to go, but you can do this as many times as you like. So, after 1 step you're half-way there. After the second, you're 3/4 there and so on. Thinking carefully about this, you will get closer and closer to the other side of the room but even after as many steps as you can take, you'll never actually reach the other side even though you're travelling further and further.

Imagine the function f(x) as a black box which takes in any value x and produces an output y. The black box acts on its input according to a rule which produces a unique value of y for a given x. If the black box becomes f(x-a) then whenever we throw a value into it a is subtracted from this value before the usual rule is applied. Thus if we imagine a plot of the function we see that any given point on the x-axis will become associated with the y-value originally paired with an adjacent point on the x-axis separated by a distance a.
For example, if we throw 4 into f(x-2)=(x-2)^2 we get 4. But we also get 4 if we throw 2 into f(x)=x^2. Similarly, we get the same output from inputting 2 into f(x-2) as from inputting 0 into f(x) and so on. Drawing an example like f(x)=x^2 and trying some example inputs and outputs should allow you to visualise how this will shift the plot depending on the value of the constant a.

Answered by James H.

Studies MSc Economics at Warwick

y = 4x^3 - 5/x^2
Easier to differentiate the 2nd term if it isn't written as a fraction so first rewrite y:
y = 4x^3 - 5x^(-2)
Then differentiate each term by multiplying each term by the current power of x and then decreasing the power of x by 1 to get
dy/dx = (4)(3)x^2 + (-5)(-2)x^(-3) = 12x^2 + 10x^(-3)

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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.

With one-to-one online support from a Maths tutor, you can spend time focusing on the topics you want to work on.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.