Currently unavailable: for regular students
Degree: Mathematics (Masters) - Oxford, Brasenose College University
Hi, I'm Oli, and I'm a Maths student at the University of Oxford - yes I really do love Maths that much! I have always enjoyed passing on my knowledge and enthusiasm of the subject to others, and have experience of tutoring several of my school friends at various stages of school-level maths.
I excelled at A-level and GCSE, so would be more than comfortable tackling difficulties in Maths and Further Maths at both of these levels, just let me know the exam-board and particular areas of difficulty and I will be able to help you to the best of my ability. (I am also willing to help with tuition for the Oxford MAT, for those who want it.)
Seeing a student of mine make a breakthrough in understanding a particular concept is what drives me on. I will be patient, persistent and will try to approach problems from various angles (badum-tsss!). I am aware that many people are not the biggest fans of Maths to say the least, but I will try my very best to make the subject more enjoyable. I will be encouraging, supportive and will try to instil confidence in my students, because it is so important that they believe they can do it!
Please don't hesitate to contact me, and I look forward to working with you!
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|Further Mathematics||IB||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|.MAT.||Uni Admissions Test||£25 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Here we need to use the chain rule because we have a function (natural log) of another function (x^2+3x+5). Let u=x^2+3x+5, and differentiate lnu with respect to u, this gives us 1/u. Then we differentiate x^2+3x+5 with respect to x, so we get 2x+3. Now the chain rule says: dy/dx=dy/du*du/dx, so we have dy/dx = (1/u)*(2x+3)=(2x+3)/(x^2+3x+5)see more