Aurimas S.

Aurimas S.

£18 - £25 /hr

Biomedical and information engineering MEng (Integrated Masters) - Cambridge University

3 completed lessons

About me

Hello everybody!

My name is Aurimas and I am currently

studying biomedical and information engineering at the University of Cambridge. I am really excited about bridging the gap between modern technologies and the medical field to provide the best healthcare for everyone.

Learning new concepts has always excited me and tutoring provides me the opportunity to spread and ignite the passion

for discovery and education in others.

Having been a mentor in a Lithuanian Oxbridge mentorship programme for three years, I have gathered extensive experience in preparing my tutees for examinations and university entry assessments/interviews by organising personal statement workshops and multiple mock interviews with the students.

Hello everybody!

My name is Aurimas and I am currently

studying biomedical and information engineering at the University of Cambridge. I am really excited about bridging the gap between modern technologies and the medical field to provide the best healthcare for everyone.

Learning new concepts has always excited me and tutoring provides me the opportunity to spread and ignite the passion

for discovery and education in others.

Having been a mentor in a Lithuanian Oxbridge mentorship programme for three years, I have gathered extensive experience in preparing my tutees for examinations and university entry assessments/interviews by organising personal statement workshops and multiple mock interviews with the students.

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About my sessions

I believe that every person is different, has different strengths and weaknesses therefore I like to tailor the style of tutoring that's best suited to my tutees - it can range from extensively covering the whole course of specific subject or identifying specific topics and exploring them further.

Working together and seeing my tutees experience joy of understanding and learning new concepts is one of the most rewarding reasons why I myself and many others choose to tutor!

I believe that every person is different, has different strengths and weaknesses therefore I like to tailor the style of tutoring that's best suited to my tutees - it can range from extensively covering the whole course of specific subject or identifying specific topics and exploring them further.

Working together and seeing my tutees experience joy of understanding and learning new concepts is one of the most rewarding reasons why I myself and many others choose to tutor!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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RR

Rodrigo Student

18 Mar

Qualifications

SubjectQualificationGrade
PhysicsOther100%
MathematicsOther100%
English LanguageOther100%
BiologyOther96%
Lithuanian language and literatureOther86%
PhysicsOther100%

General Availability

MonTueWedThuFriSatSun
Pre 12pm
12 - 5pm
After 5pm

Pre 12pm

12 - 5pm

After 5pm
Mon
Tue
Wed
Thu
Fri
Sat
Sun

Subjects offered

SubjectQualificationPrice
MathsIB£22 /hr
PhysicsIB£22 /hr
Personal StatementsMentoring£22 /hr
MathsUniversity£25 /hr
PhysicsUniversity£25 /hr

Questions Aurimas has answered

Find the derivative of a function y(x) = x^x

It is easy to see that this function can't be explicitly differentiated so we shall use implicit differentiation in this case. By applying natural log to both sides we obtain:ln(y(x)) = ln(xx) = xln(x), where ln(x) - logarithm of x with base e. Now we can differentiate both sides with respect to x.In the left hand side we obtain dln(y(x))/dx = (1/y(x)) * y'(x) = y'(x)/(xx) [Chain rule]In the right hand side we obtain d(xln(x)) = ln(x) + x/x = ln(x) + 1 [Product rule]Finally, we equate both sides and get y'(x)/(xx) = ln(x) + 1, and by multiplying both sides by xx we obtain y'(x) = (xx)*(ln(x) + 1)It is easy to see that this function can't be explicitly differentiated so we shall use implicit differentiation in this case. By applying natural log to both sides we obtain:ln(y(x)) = ln(xx) = xln(x), where ln(x) - logarithm of x with base e. Now we can differentiate both sides with respect to x.In the left hand side we obtain dln(y(x))/dx = (1/y(x)) * y'(x) = y'(x)/(xx) [Chain rule]In the right hand side we obtain d(xln(x)) = ln(x) + x/x = ln(x) + 1 [Product rule]Finally, we equate both sides and get y'(x)/(xx) = ln(x) + 1, and by multiplying both sides by xx we obtain y'(x) = (xx)*(ln(x) + 1)

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1 month ago

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