Currently unavailable: for new students
Degree: MMath ( G103) (Masters) - Warwick University
I am a cheerful experienced tutor, having taught both in classrooms and one-on-one. Doing so has taught me a range of teaching techniques such that if one doesn't work for you, you can rest assured I have one that will help.
I have always had a thirst for knowledge and if you too have a thirst, should you have any questions I would welcome them and answer them to the best of my capabilities. I have taught a large volume of students with a scope of abilities and so would be able to help even the most able minded students.
Exam Boards: Maths/ Further Maths - Edexcel, Physics - OCR
Availability: I'm free most days past 5 and on the weekends. My timetable changes often, so just ask and i'll try and let you know!
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|.STEP.||Uni Admissions Test||£26 /hr|
|STEP I||Uni Admissions Test||1|
|STEP II||Uni Admissions Test||3|
Prince (Student) October 29 2015
John (Student) October 8 2015
John (Student) October 7 2015
John (Student) September 17 2015
y = x^x -- Yikes! Doesn't that look ugly.
It's mostly the x^x part, since ... to the x is fine, and x to the ... is fine.
We must split the two x's in order to continue.
Perhaps we could log? since then we can pull the index to the front.
so: log(y) = log(x^x) = x log(x) -- because [log(a^b) = b log(a)]
Well this looks much better. On the left we have something which is easy using chain rule
and the right side, looks easy using product rule.
d/dx log(y) = dy/dx * (1/y)
d/dx x*log(x) = 1*log(x) + x*(1/x)
= log(x) + 1
dy/dx * (1/y) = 1 + log(x)
Multiplying through by y, gives us:
dy/dx = y ( 1 + log(x) )
but remember, y = x^x
so dy/dx = x^x ( 1 + log(x) )