Pascal L. GCSE Maths tutor, A Level Maths tutor, GCSE Biology tutor, ...

Pascal L.

Currently unavailable: for new students

Studying: Engineering (Masters) - Durham University

5.0
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1 review| 5 completed tutorials

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About me

I’m a tutor from St. Aiden’s College, Durham, studying an integrated masters in engineering. Before going to university I was a Young Leader with the Scout Association, and a member of the Air Training Corps - I believe the skills I've picked as a member of these organisations was why I was picked to be a sixth form maths buddy, tutoring GCSE maths students who were struggling with the course.  I was also a member  of the Northumbrian Univeristy Officer Training Corps (NUOTC) for a year before deciding to complete the reserve commisioning course over the summer of 2015, after which I transferred to a Royal Engineers unit. A large part of the training I have attended with the military links directly to leadership and time management.

I studied AQA chemistry A level and science GCSE’s, Edexcel maths A level, further maths(AS) and biology (AS) along with OCR’s Advancing Physics (B) A level course and a Free Standing Maths Qualification (FSMQ). I tutor science and maths GCSE/A levels, but am also ready to help students produce fluid, engaging personal statements.  

I’m generally available weekday afternoons and evenings (except Teusdays), but if you’re looking for something outside those times (or non-regular tutorials) drop me a message and I’ll get back to you, or mention it in a ‘meet the tutor’ video call!

I’m a tutor from St. Aiden’s College, Durham, studying an integrated masters in engineering. Before going to university I was a Young Leader with the Scout Association, and a member of the Air Training Corps - I believe the skills I've picked as a member of these organisations was why I was picked to be a sixth form maths buddy, tutoring GCSE maths students who were struggling with the course.  I was also a member  of the Northumbrian Univeristy Officer Training Corps (NUOTC) for a year before deciding to complete the reserve commisioning course over the summer of 2015, after which I transferred to a Royal Engineers unit. A large part of the training I have attended with the military links directly to leadership and time management.

I studied AQA chemistry A level and science GCSE’s, Edexcel maths A level, further maths(AS) and biology (AS) along with OCR’s Advancing Physics (B) A level course and a Free Standing Maths Qualification (FSMQ). I tutor science and maths GCSE/A levels, but am also ready to help students produce fluid, engaging personal statements.  

I’m generally available weekday afternoons and evenings (except Teusdays), but if you’re looking for something outside those times (or non-regular tutorials) drop me a message and I’ll get back to you, or mention it in a ‘meet the tutor’ video call!

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10/12/2013

Ratings & Reviews

5from 1 customer review
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Georgie (Student)

April 29 2015

Really useful session going over past papers.

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
ChemistryA-level (A2)A*
PhysicsA-level (A2)A*
BiologyA-level (AS)A
Further MathsA-level (AS)A

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
PhysicsA Level£20 /hr
BiologyGCSE£18 /hr
ChemistryGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
ScienceGCSE£18 /hr

Questions Pascal has answered

How do I differentiate y=x^x?

y=x^x

To find the differential, dy/dx, logs of both sides must be taken:

log(y)=log(x^x)

Then using log rules, the power can be brought down, outside the log expression:

log(y) = x log(x)

This expressions can now be differentiated with respect to x, using the chain rule on the left and the product rule on the right, giving:

(1/y) * dy/dx = 1 + log(x)

Multiplying through by y gives:

dy/dx = y (1 + log(x)) 

Remember! From the start of the question y=x^x, so this can be rewritten to:

dy/dx = x^x + x^xlog(x)

 

y=x^x

To find the differential, dy/dx, logs of both sides must be taken:

log(y)=log(x^x)

Then using log rules, the power can be brought down, outside the log expression:

log(y) = x log(x)

This expressions can now be differentiated with respect to x, using the chain rule on the left and the product rule on the right, giving:

(1/y) * dy/dx = 1 + log(x)

Multiplying through by y gives:

dy/dx = y (1 + log(x)) 

Remember! From the start of the question y=x^x, so this can be rewritten to:

dy/dx = x^x + x^xlog(x)

 

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3 years ago

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