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About me

Hi there!

To kick off, a little about my backround. I recently graduated with a B.A. Hons in Mathematics from the University of Cambridge, and am about to start an MSc at Imperial College London. Prior to university, I achieved A*s in A-Level Mathematics, Further Mathematics (Edexcel), Physics (AQA), and History (Edexcel). 

I have now been an active on MyTutor for just under two years, and have tutored over 25 students, varying from those just starting out on their GCSEs, to prospective Oxbridge students looking for guidance with their entrance exams. Aside from online tutoring, I have also gained invaluable experience during in intensive four week programme in June-July 2014 teaching (English) in a school in South Korea, and volunteering as an assistant mathematics teacher at a local secondary school during my time at university. 

I have a very flexible approach to tutoring, and am very happy to adapt my approach to the requirements of any student. In general, I strongly believe that a discursive and interactive approach to teaching and learning is the most effective and enjoyable. Above all, students do their best when they enjoy what they are doing!

Please get in touch if you would like to discuss your tutoring requirements further! 

Subjects offered

SubjectLevelMy prices
Extended Project Qualification A Level £30 /hr
Further Mathematics A Level £30 /hr
Further Mathematics A Level £30 /hr
Maths A Level £30 /hr
Maths A Level £30 /hr
Physics A Level £30 /hr
Further Mathematics GCSE £30 /hr
Maths GCSE £30 /hr
Maths GCSE £30 /hr
Physics GCSE £30 /hr
Physics GCSE £30 /hr
-Oxbridge Preparation- Mentoring £30 /hr
-Personal Statements- Mentoring £30 /hr
.MAT. Uni Admissions Test £30 /hr
.STEP. Uni Admissions Test £30 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
HistoryA-LevelA*
STEP IIUni Admissions Test2
STEP IIIUni Admissions Test1
French (AS)A-LevelA
MathematicsBachelors Degree2.1
Advanced Extension Award (Mathematics)A-LevelDistinction
Disclosure and Barring Service

CRB/DBS Standard

11/11/2014

CRB/DBS Enhanced

No

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

5from 49 customer reviews

Yuan (Student) October 18 2016

very good tutor

Maryam (Parent) September 18 2016

Another excellent tutorial

Sheila (Parent) September 13 2016

Brilliant, thanks for the last minute review for a quiz. It really saved me!

Sheila (Parent) September 13 2016

Great!
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Questions Louis has answered

How do you differentiate x^x?

There are two ways we can find the derivative of x^x. It's important to notice that this function is neither a power function of the form x^k nor an exponential function of the form b^x, so we can't use the differentiation formulas for either of these cases directly. (i) Let y=x^x, and take l...

There are two ways we can find the derivative of x^x. It's important to notice that this function is neither a power function of the form x^k nor an exponential function of the form b^x, so we can't use the differentiation formulas for either of these cases directly.

(i) Let y=x^x, and take logarithms of both sides of this equation: ln(y)=ln(x^x). Using properties of logarithmic functions, we can rewrite this as ln(y)=x.ln(x). Then differentiating both sides with respect to x, and using the chain rule on the LHS and product rule on the RHS, gives 1/y.dy/dx=ln(x)+1. Rearranging, we have dy/dx=y.(ln(x)+1). That is, dy/dx=x^x(ln(x)+1).

(ii) Write x^x=e^(ln(x^x))=e^(x.ln(x)), using the properties of the exponential and logarithmic functions. Now, d/dx(x.ln(x))=ln(x)+1 by the product rule. Hence, d/dx(e^(x.ln(x)))=(ln(x)+1).(e^(x.ln(x)) by the chain rule, and using the fact that the derivative of e^[f(x)]=f'(x).e^[f(x)] for any differentiable function f(x). Finally, rewriting e^(x.ln(x)) as x^x gives d/dx(x^x)=x^x.(ln(x)+1), as with the first method.

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2 years ago

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