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 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.

Problems of this style are solved using the chain rule.

To begin, define the quantity inside the brackets as *u*

**u = 6x-1** such that **y = u^7**

It is now useful to write the chain rule. We can see

**dy/dx = du/dx x dy/du**

as the ** du **'cancel'. Now, all we need to do is differentiate two simple expressions:

**du/dx = 6 **and **dy/du = 7u^6**

Substituting these expressions back into the chain rule:

**dy/dx = 42u^6**

Finally, substitute ** u **into this expression to give the final answer,

**dy/dx = 42(6x-1)^6**

When proving trigonometric identites, we must show that the left hand side of the equation = the right hand side. Here we will start with the left hand side (LHS) and show that it is equivalent to the right hand side (RHS).

LHS=2sin(2x)-3cos(2x)-3sin(x)+3

Using the double angle rules for sin(2x) and cos(2x);

LHS=2(2sin(x)cos(x))-3(cos^{2}(x)-sin^{2}(x))-3sin(x)+3

Notice that the RHS has sin(x) factorised out, meaning that every term in the LHS has a common factor of sin(x). Currently the LHS has a cos^{2}x term, but we can change this to a sin^{2}x term using the identity: cos^{2}(x)=1-sin^{2}(x)

LHS=2(2cos(x)sin(x))-3(1-sin^{2}(x)-sin^{2}(x))-3sin(x)+3

=4cos(x)sin(x)-3(1-2sin^{2}(x))-3sin(x)+3

=4cos(x)sin(x)-3+6sin^{2}(x)-3sin(x)+3

=4cos(x)sin(x)+6sin^{2}(x)-3sin(x)

=sin(x)(4cos(x)+6sin(x)-3)

=RHS

We have shown that LHS=RHS, therefore the proof is complete.

The method we use to differentiate this form of equation is called the chain rule.

The chain rule is dy/dx = dy/du x du/dx

We can rememeber the right way up of the terms on the right hand side by treating them as fracions and cancelling to give dy/dx.

To use the chain rule we need to define our u. In this form of question we choose what is inside the brackets.

Let u=4+9x, this means that y=u^5.

Then by normal rules of differentiation we differentiate y and u giving:

dy/du = 5u^4 and du/dx = 9

Then we substitue these results into the chain rule formula giving:

dy/dx = 9 x 5u^4 = 45u^4

Then we substitute u=4+9x back in to get our final answer:

dy/dx = 45(4+9x)^4

Answered by Jenny H.

Studies Economics and Mathematics at Bristol

This is a quadratic equation and as such it has zero, one or two solutions depending on the value of the discriminant (b^{2}-4ac). In this equation, a=3, b=-6 and c=2 so b^{2}-4ac = 36-24=12. As this is >0 the equation has two real solutions, however this is not a square number and therefore we cannot factorise and will have to use the quadratic formula. This is (-b (+/-) (b^{2}-4ac)^{1/2})/(2a). Subsituting in a, b and c gives us (6 (+/-) 12^{1/2})/6 which means our two solutions are x=1+(1/6)12^{1/2}and x=1-(1/6)12^{1/2}

Answered by Angus S.

Studies Cognitive Science (BSc Hons) at Edinburgh

The easiest way to see this is with an example. So let us take the curve:

x = cos(2t) , y = sin(t) with a point P where t = pi/6

The gradient at any point is equal to the derivative of y with respect to x at the given point. But in this situation we cannot simply take dy/dx. However under closer inspection and treating dy/dx as a fraction we see that dy/dx = (dy/dt)(dt/dx). However, this dt/dx may also present problems, but it also can be seen as 1/(dx/dt) in treating it as a fraction as well.

So all we need to calculate is dy/dt and dx/dt:

dy/dt = cos(t) , dx/dt = -2sin(2t)

Now dt/dx = 1/(dx/dt) = -1/(2sin(2t))

Then dy/dx = (dy/dt)(dt/dx) = -cos(t)/2sin(2t)

So we now have the derivative at all points in the curve. To find the gradient at the point P we simply substitute in the value of t and calculate.

This gives:

-cos(pi/6)/2sin(pi/3) = -(3^{1/2}/2)/2(3^{1/2}/2) = -1/2

Thus the gradient of this parametric equation at P is -1/2. This method can easily be extended and adapted to all equations of parametric form.

Answered by Alex D.

Studies Maths and Philosophy at Bristol

You can express tan(x) as sin(x)/cos(x).

Therefore, tan(x)= sin(x)/ cos(x)

The quotient rule can be applied here as there is a function of x in the numerator and denominator.

Quotient Rule: (v*(du/dx) - u*(dv/dx))/v^{2}

Let u =sin(x) and v=cos(x) and hence (du/dx)= cos(x) and (dv/dx)= -sin(x).

Therefore:

d(tan(x))/dx= (cos(x)*cos(x))-(sin(x)*(-sin(x))/(cos^{2}(x))

=(cos^{2}(x)+sin^{2}(x))/(cos^{2}(x))

Using the trig identity, cos^{2}(x)+sin^{2}(x)=1, the numerator of the fraction can be tidied and heavily simplified.

d(tan(x))/dx= 1/(cos^{2}(x))

As 1/(cos(x)) is equal to sec(x), 1/(cos^{2}(x)) is equal to sec^{2}(x).

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If you're taking A-Level Maths - well done to you. Its one of the trickier A-levels but also one of the best respected, and it will set you up well for all sorts of careerÂ paths. That said, it can be a surprising jump up from the Maths you've done before. And because the course progresses so quickly, its easy to feel like you're getting left behind.

With one-to-one support from a Maths tutor you can spend time focusing on the topics you want to work on.Whether its 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 feel confident, well prepared, and able to achieve the results you want.