Ben B. A Level Maths tutor, A Level Further Mathematics  tutor, A Lev...

Ben B.

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Mathematics and Physics (Masters) - Bristol University

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About me

Hi. I'm Ben and I am a second year mathematician and physicist at the University of Bristol. I am looking for students to tutor for both maths and physics.

Maths and Physics are very beautiful subjects. I hope that, through my enthusiastic approach, we can develop a strong foundation in either maths or physics (or both) that will prove to be extremely useful in your studies and will help you boost your exam grades. I believe that the most important part of learning these subjects is developing a true understanding of the material. Often, a lot of the beauty and elegance of maths and physics is lost by teachers who place a huge emphasis on training students to robotically answer questions. Whilst I appreciate that answering questions and solving problems is a huge part of the learning process and will provide questions and examples in the tutorials, I would love to help you gain a deeper understanding of the theories in order for you to have a greater appreciation of the subjects. I am confident that this approach will prove extremely useful when it comes to helping you prepare for exams.

Maths units I can help you with: C1-C4, FP1-FP3, M1-M2

I look forward to meeting you!

Hi. I'm Ben and I am a second year mathematician and physicist at the University of Bristol. I am looking for students to tutor for both maths and physics.

Maths and Physics are very beautiful subjects. I hope that, through my enthusiastic approach, we can develop a strong foundation in either maths or physics (or both) that will prove to be extremely useful in your studies and will help you boost your exam grades. I believe that the most important part of learning these subjects is developing a true understanding of the material. Often, a lot of the beauty and elegance of maths and physics is lost by teachers who place a huge emphasis on training students to robotically answer questions. Whilst I appreciate that answering questions and solving problems is a huge part of the learning process and will provide questions and examples in the tutorials, I would love to help you gain a deeper understanding of the theories in order for you to have a greater appreciation of the subjects. I am confident that this approach will prove extremely useful when it comes to helping you prepare for exams.

Maths units I can help you with: C1-C4, FP1-FP3, M1-M2

I look forward to meeting you!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
BiologyA-level (A2)A

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
PhysicsA Level£20 /hr

Questions Ben has answered

differentiate y = (4-x)^2

This is a basic example of a very important result: the chain rule. The difficulty of this sort of example is that we have a "function of a function". That is, we have the function '4-x' and then we square this. 

The general approach is as follows. First we will let 'u' be a new function: u=4-x. It is evident that now we have y=u^2 which looks like it might be easier to work with. The chain rule says the following:

dy/dx = (dy/du)*(du/dx)

In this case y=u^2 so, from normal differentiation, we get dy/du = 2u. We also then have u = 4-x. So, again from normal differentiation techniques, we have du/dx=-1.

Using the chain rule gives dy/dx = (2u)*(-1) and if we substitute u=4-x we get

dy/dx = -2(4-x)  = 2x-8    which is the final answer.                                   

This is a basic example of a very important result: the chain rule. The difficulty of this sort of example is that we have a "function of a function". That is, we have the function '4-x' and then we square this. 

The general approach is as follows. First we will let 'u' be a new function: u=4-x. It is evident that now we have y=u^2 which looks like it might be easier to work with. The chain rule says the following:

dy/dx = (dy/du)*(du/dx)

In this case y=u^2 so, from normal differentiation, we get dy/du = 2u. We also then have u = 4-x. So, again from normal differentiation techniques, we have du/dx=-1.

Using the chain rule gives dy/dx = (2u)*(-1) and if we substitute u=4-x we get

dy/dx = -2(4-x)  = 2x-8    which is the final answer.                                   

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2 years ago

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