PremiumRobbie H. A Level Maths tutor, GCSE Maths tutor, 11 Plus Maths tutor,...

Robbie H.

Currently unavailable: for regular students

Degree: Engineering Design (Masters) - Bristol University

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About me

I am a 22 year old Master's student at the University of Bristol studying Engineering Design. I achieved a First in my Bachelors and A*A*A at A Level. I have a passion for maths and science and hope that in my sessions some of this will rub off on you!

I am friendly, approachable and very patient! I am new to this website, but not to tutoring. This is my third year working as a tutor. For the academic year 14/15, I tutored three students (Year 8, two Year 12s) and in 15/16, I tutored five students (Year 9, two Year 11s, two Year 13s). So far every student I've tutored has met or exceeded the grade they were aiming for! I received a terrific response from all students and thoroughly enjoyed seeing their confidence and ability in maths grow. All three students from 14/15 kept me as a tutor for 15/16.

On top of this, I am on the committee of Engineers Without Borders as the Outreach Officer. I regularly attend local schools of all ages, where I lead workshops raising awareness about science and the environment. This has led to me teaching children a wide range of topics under the STEM umbrella. As a student myself, I understand the stress of exams and like to see myself as the student's peer, listening to their feedback and tailoring the sessions to their needs.

The Sessions:

Understanding is key in maths and science and therefore I tend to focus on helping the student to get to grips with what can be at first challenging topics. I personally found the best way to learn was to do a bit of theory followed by loads of examples. However, if the student does not respond brilliantly to this method, I can easily adapt to suit their needs.

What Next?

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website). Remember to tell me your exam board and what you're struggling with.

I look forward to meeting you!

Subjects offered

SubjectQualificationPrices
Maths A Level £26 /hr
Maths GCSE £24 /hr
Physics GCSE £24 /hr
Maths 13 Plus £24 /hr
Maths 11 Plus £24 /hr

Qualifications

SubjectQualificationLevelGrade
Engineering DesignDegree (Bachelors)First
MathsA-levelA2A*
PhysicsA-levelA2A
BiologyA-levelA2A*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

Currently unavailable: for regular students

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Questions Robbie has answered

Solve the equation d/dx((x^3 + 3x^2)ln(x)) = 2x^2 + 5x, leaving your answer as an exact value of x. [6 marks]

This may look complicated at first but it can be broken down into a number of simple steps... For the left side of the equation, the d/dx tells you we are differentiating and since (x3 + 3x2)ln(x) sees two functions multiplied together, we therefore use the product rule. The product rule loo...

This may look complicated at first but it can be broken down into a number of simple steps...

For the left side of the equation, the d/dx tells you we are differentiating and since (x3 + 3x2)ln(x) sees two functions multiplied together, we therefore use the product rule.

The product rule looks like this:

u v' + u' v

where in this case u = (x3 + 3x2) and v = ln(x)

u' is the differential of u. Differentiating here simply involves multiplying the coefficient (number in front of the x) by the power and then subtracting one from the power so:

 u' = 3x2 + 6x

v' is the differential of v so:

v' = 1/x

Now we simply put the values into our product rule equation and put into our product rule equation:

u v' + u v'

(x3 + 3x2)(1/x) + (3x2 + 6x)(ln(x))

Expanding the first set of brackets gives:

x2 + 3x + (3x2 + 6x)(ln(x))

Putting this into the equation stated in the question gives:

x2 + 3x + (3x2 + 6x)(ln(x)) = 2x2 + 5x

Rearrange and simplify as follows:

(3x2 + 6x)(ln(x)) = x2 + 2x

ln(x) = (x2 + 2x) / (3x2 + 6x)

Now if we factorise the bottom of the fraction:

ln(x) = (x2 + 2x) / (2(x2 + 3x))

And cancelling the x2 + 2x terms:

ln(x) = 1/3

Solving for x gives:

x = e(1/3)

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8 months ago

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