Sam E. A Level Maths tutor, GCSE Maths tutor

Sam E.

Unavailable

Computer Science (Masters) - York University

4.8
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4 reviews

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

6 completed lessons

About me

About me:

Hello! I'm a first year Computer Science student at the University of York.

I offer tutoring for maths at GCSE and A-Level. Despite not studying maths now, it was my original love - the subject which first taught me how to enjoy solving the even the most devious of problems. Through my tutoring sessions, I hope to inspire some of the same enthusiasm for maths in my students. 

Tutoring sessions:

The tutoring sessions are quite flexible and can be changed for the tutee's needs. I am happy to work through a problem with them that they are stuck on. Alternatively, the tutee can suggest a topic they want practice in, and I can work through common questions with them.

Contact me: 

Feel free to send me a message! Tell me what topic/question you're struggling with, and it would also be helpful if you tell me what exam board you study. 

About me:

Hello! I'm a first year Computer Science student at the University of York.

I offer tutoring for maths at GCSE and A-Level. Despite not studying maths now, it was my original love - the subject which first taught me how to enjoy solving the even the most devious of problems. Through my tutoring sessions, I hope to inspire some of the same enthusiasm for maths in my students. 

Tutoring sessions:

The tutoring sessions are quite flexible and can be changed for the tutee's needs. I am happy to work through a problem with them that they are stuck on. Alternatively, the tutee can suggest a topic they want practice in, and I can work through common questions with them.

Contact me: 

Feel free to send me a message! Tell me what topic/question you're struggling with, and it would also be helpful if you tell me what exam board you study. 

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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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Ratings & Reviews

4.8from 4 customer reviews
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Barbara (Parent from Fareham)

January 25 2017

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Abbie Hollis (Student)

January 13 2017

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Abbie Hollis (Student)

January 5 2017

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Barbara (Parent from Fareham)

December 2 2016

Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A*
PhysicsA-level (A2)A*
ComputingA-level (A2)A*

General Availability

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

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions Sam has answered

How would you differentiate f(x) = 2x(3x - 1)^2 using the chain rule?

In order to differentiate this expression, you need to use the chain rule. 

The chain rule gives: f'(x) = uv' + u'v.

The u and the v are two parts of the original function f(x). The apostrophe ' at the end means the derivative of that.

We need to assign values to u and v, so we look at the function f(x) = 2x(3x - 1)2 to see what parts it is in:

u = 2x

v = (3x - 1)2

Then, we differentiate each of these.

u' = 2

v' = 2 x 3 x (3x - 1)1 = 6(3x - 1)

Now, we can put this expression altogether:

f'(x) = uv' + u'v = 2x(6(3x - 1)) + 2(3x - 1)2

And now, simplify.

f'(x) = 12x(3x - 1) + 2(3x - 1)2

f'(x) = 2(3x - 1)[6x + (3x - 1)]

f'(x) = 2(3x - 1)(9x - 1)

f'(x) = 2(27x2 - 9x - 3x + 1)

f'(x) = 2(27x2 - 12x + 1)

f'(x) = 54x2 - 24x + 2

In order to differentiate this expression, you need to use the chain rule. 

The chain rule gives: f'(x) = uv' + u'v.

The u and the v are two parts of the original function f(x). The apostrophe ' at the end means the derivative of that.

We need to assign values to u and v, so we look at the function f(x) = 2x(3x - 1)2 to see what parts it is in:

u = 2x

v = (3x - 1)2

Then, we differentiate each of these.

u' = 2

v' = 2 x 3 x (3x - 1)1 = 6(3x - 1)

Now, we can put this expression altogether:

f'(x) = uv' + u'v = 2x(6(3x - 1)) + 2(3x - 1)2

And now, simplify.

f'(x) = 12x(3x - 1) + 2(3x - 1)2

f'(x) = 2(3x - 1)[6x + (3x - 1)]

f'(x) = 2(3x - 1)(9x - 1)

f'(x) = 2(27x2 - 9x - 3x + 1)

f'(x) = 2(27x2 - 12x + 1)

f'(x) = 54x2 - 24x + 2

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2 years ago

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