PremiumMayur D. A Level Maths tutor, A Level Further Mathematics  tutor, GCS...

Mayur D.

Currently unavailable: until 30/06/2017

Degree: MMath ( G103) (Masters) - Warwick University

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About me

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!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £24 /hr
Maths A Level £24 /hr
Maths GCSE £22 /hr
Physics GCSE £22 /hr
.STEP. Uni Admissions Test £26 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA
STEP IUni Admissions Test1
STEP IIUni Admissions Test3
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

30/06/2017

Ratings and reviews

4.9from 21 customer reviews

Prince (Student) October 29 2015

Absolutely top notch tutoring, would fully recommend!

John (Student) October 8 2015

great

John (Student) October 7 2015

Great

John (Student) September 17 2015

Amazing
See all reviews

Questions Mayur has answered

Find the derivative of x^x

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) ...

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.

So:
d/dx log(y) = dy/dx * (1/y) 
and:
d/dx x*log(x) = 1*log(x) + x*(1/x)
= log(x) + 1

so:
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) )

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2 years ago

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