I cannot praise this tutor enough. Highly skilled in mathematics, teaches at the student's pace, patient and super efficient. A conscientious person who responds immediately to messages. The lesson was tailor made to suit my own needs. Mayur is excellent at pin pointing math problems quickly - taking things back to basic principals where needed. A most enjoyable tutorial - I would highly recommend this very gifted tutor to students of every ability.

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Parent review - it's a relief to see my son start to understand GCSE maths topics that he has struggled with. His confidence is growing with Chris's help. Chris isn't phased by any questions or topics thrown at him, and is able to clearly explain the methods needed to answer maths questions.

Kelly, Parent from Kent

Why waste time looking locally when itâ€™s easier to find the right tutor online?

Tutoring is easier and more flexible when you remove the need to plan around travel

With live online one-to-one sessions you’re always engaged

Your live sessions are recorded, so you can play tutorials back if you want to revise

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.

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).

Answered by Chinazam U.

Studies MEng in Materials Engineering with International Study at Exeter

Begin by adding the 2 ratio parts (i.e. the numbers which make up the ratio) so 8+5 = 13.

Then divide the total sum by the sum of the ratio parts. Thus £650/13 = £50.

Multiply this by each of the ratio parts.

£50 * 8 = £400

£50 * 5 = £250

Finish by returning the answers to the correct form. Thus the answer is £400:£250

Answered by Harry V.

Studies Politics and Economics at Exeter

At first glance this seems as though we need to solve using the techniques of standard differentiation, however on further inspection we see we need to use a further method call the chain rule to solve this.

The chain rule uses the idea of dy/dx = dy/du X du/dx

(a way to remember to get the fractions the right way up on the right hand side, is to treat the entities as regular fractions and cancelling should leave dy/dx)

To use the chain rule substitute u = 3 + 6x

So y = u^{5}

dy/du = 5u^{4} and du/dx = 6 (differentiate both equations)

sub these two differentials into dy/dx = dy/du X du/dx

So dy/dx = 5u^{4} X 6 = 30u^{4}

Now substitute u = 3 + 6x back in

dy/dx = 30(3 + 6x)^{4}

Answered by Tom C.

Studies Engineering Science at Oxford, Jesus College

So x^{2 }+ 9x + 20 = 0

My preffered way of solving this equation is to factorise the equation. (Though I understand that different students may find other ways easier)

Factorisation is where the above equation is (x+a)(x+b) = 0

So if we times out (x+a)(x+b) we get

x^{2} + ax + bx + ab = 0

therefore

x^{2} + (a+b)x + ab = 0

Therefore we can equate this to the original question, so

x^{2 }+ 9x + 20 = x^{2} + (a+b)x + ab

so now we can see that

9 = a + b and

20 = ab

I would reccomend using trial and error (although I understand that different students may prefer other techniques).

So by trying for multiple values of a and b, we can see that they must equal 5 and 4.

Therefore

x^{2} + 9x + 20 = (x+5)(x+4) = 0

We know that the only way of producing a 0 through multiplication is through multiplying one number by another. Therefore we know that

x+5= 0 or x+4=0

Through rearranging these equations we can conclude that x must equal -4 or -5.

Answered by Tilly P.

Studies Computer Science and Mathematics at Exeter

Let's draw a line. It crosses the point (0,1), intercepting the y-axis (remember, y to the sky!) at 1 and the x-axis at the point (-2,0).

An equation tells us what a line looks like, it's a way of describing a line in a mathematical way. If we do it right, we can draw a line just by looking at the equation. Equations for straight lines are ususally given in the form y=mx+c - this tells us what y values do for different values of x. But what are m and c?

So, in order to find the equation for any straight line, we need two things: the gradient and the y-intercept. We call the y-intercept c and the gradient m. As soon as we know these two things, the rest is easy!

First step: finding the y-intercept - this is the simple bit, all you need to do is look at the y-axis and see where the line crosses it. In this case it is 1, so c=1.

Second step: finding the gradient - this requires a little bit more thought. All gradient means is the steepness of the line, so for every place the line moves to the right, how many places up does it go? In this case the line is not very steep at all and, in fact, you can see that for every place you move to the right it only goes up half a place up. Therefore m=0.5.

The tricky bit is over! Now comes the simple bit. All you have to do now is sub the values you've found for m and c back into the template equation, y=mx+c. So, if c=1 and m=0.5 then y=0.5x+1 - problem solved!

Answered by Dominique G.

Studies Classics at Cambridge

let y=sec(x) = 1/(cos(X)) = cos(x)^{-1}

Thus dy/dx = -1(cos(x))^{-2}(-sinx) = sin(x)/(cos(x))^{2}

= 1/cos(x) x sin(x)/cos(x)

=sec(x)tan(x)

Answered by Owain D.

Studies Mathematics and Economics at Bristol

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