Elizabeth H. A Level Chemistry tutor, GCSE Chemistry tutor, A Level P...

Elizabeth H.

Currently unavailable: for regular students

Degree: Natural Science (Masters) - Durham University

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About me

Hi! I'm a Natural Sciences student at Durham University studying Physics and Chemistry. I love learning science and applying it to the real world as well as exam and homework questions, and would love to help you do the same!

I tutored Chemistry and Physics while I was at college, so have lots of experience in finding excatly what will help you. I'm friendly and approachable, and having taken time to understand new concepts myself I know that no question is a stupid one!

In our sessions you will get the chance to ask me lots of questions and gain understanding of everything you've learnt. We will cover which ever topics you want - just tell me which and your exam board if applicable. 

To get in touch, please contact me via 'WebMail' or a 'Meet a Tutor Session'. I look forward to seeing you!

Subjects offered

SubjectLevelMy prices
Chemistry A Level £22 /hr
Further Mathematics A Level £22 /hr
Maths A Level £22 /hr
Physics A Level £22 /hr
Chemistry GCSE £20 /hr
Further Mathematics GCSE £20 /hr
Maths GCSE £20 /hr
Physics GCSE £20 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
Chemistry A-LevelA*
PhysicsA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Ratings and reviews

5from 8 customer reviews

Mya (Student) November 12 2016

Mya (Student) August 29 2016

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Questions Elizabeth has answered

A curve has the equation y = x^4 - 8x^2 + 60x + 7. What is the gradient of the curve when x = 6?

To find the gradient of any curve, we take the derivative. So in this case, we need to take dy/dx. We do this by multiplying the term by the power on x, and then lowering the power by one. For example, for the first term, x4, the power is four, so we multiply x4 by four, and the power becomes ...

To find the gradient of any curve, we take the derivative. So in this case, we need to take dy/dx. We do this by multiplying the term by the power on x, and then lowering the power by one. For example, for the first term, x4, the power is four, so we multiply x4 by four, and the power becomes three, so we have 4x3. We repeat this for all of the terms individually to get dy/dx = 4x-16x +60. 

That gives us the gradient at any point. To get the gradient at x = 6 we need to substitute the value in to the new equation, so we get dy/dx = 4 * 63 - 16 * 6 + 60 = 828

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7 months ago

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