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

Mayur D.

Currently unavailable: for regular students

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

4.9
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21 reviews| 91 completed tutorials

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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!

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!

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About my sessions

I try to go through problems slowly at first to pinpoint any errors. After this, I explain the error and go through some examples slow enough to make sure there is no confusion. After this, I set homework of questions, only wanting to hear about any questions you seriously struggle on. The best way to learn any complex idea in mathematics is by practicing.

I try to go through problems slowly at first to pinpoint any errors. After this, I explain the error and go through some examples slow enough to make sure there is no confusion. After this, I set homework of questions, only wanting to hear about any questions you seriously struggle on. The best way to learn any complex idea in mathematics is by practicing.

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Ratings & Reviews

4.9from 21 customer reviews
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Prince (Student)

October 29 2015

Absolutely top notch tutoring, would fully recommend!

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John (Student)

October 8 2015

great

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John (Student)

October 7 2015

Great

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John (Student)

September 17 2015

Amazing

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A*
PhysicsA-level (A2)A
STEP IUni admission test1
STEP IIUni admission test3
MathsDegree (Masters)2:1

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£24 /hr
MathsA Level£24 /hr
MathsA Level£24 /hr
Further MathematicsGCSE£22 /hr
MathsGCSE£22 /hr
PhysicsGCSE£22 /hr
PhysicsGCSE£22 /hr
.STEP.Uni Admissions Test£26 /hr
.STEP.Uni Admissions Test£26 /hr

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) = 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) )

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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