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.
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.
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!
|Maths||A Level||£26 /hr|
|Maths||13 Plus||£24 /hr|
|Maths||11 Plus||£24 /hr|
|Engineering Design||Bachelors Degree||First|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
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)see more