Currently unavailable: for regular students
Degree: Natural Science (Masters) - Durham University
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!
|Chemistry||A Level||£22 /hr|
|Further Mathematics||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|Physics||A Level||£22 /hr|
|Further Mathematics||GCSE||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Mya (Student) December 16 2016
Mya (Student) November 12 2016
Mya (Student) August 29 2016
Mya (Student) August 16 2016
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 = 4x3 -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 = 828see more