Currently unavailable: for new students
Degree: Economics and Mathematics (Bachelors) - Bristol University
I'm currently an undergraduate at the University of Bristol studying Economics & Mathematics (Joint Honours). I am one of those few who have a genuine love of studying Maths and hope that enthusiasm will help you in tutorials and perhaps even be transferred on to you!
I left school in 2014 having achieved A*’s in both Maths and Further Maths, whilst also balancing work with the role of School Captain. I’m a confident and approachable guy who is eager to help those who I can with their understanding of Maths. I have a wealth of experience mentoring and tutoring other students during my A-Level studies, and this is where I learned first-hand that ‘doing’ is the only way to excel in Maths.
About the Tutorials:
Maths is a subject that is mastered by practicing, so my tutorials are exercise centred after some prior explanation. I will get some involvement from you so that we both ensure that you are understanding what is happening. I am very easy to talk to which is important because an open dialogue is vital for a productive tutorial; I want you to get the most out of these sessions as possible – a lot of ground can be made in 55 minutes and when you’re actively taking part then the time will fly by!
The topic of the tutorials will be guided in advance by you - I don't want to waste your time and money teaching you what you already know!
Feel free to get in touch with me via the MyTutorWeb website to schedule a session. Remember to include what you wish to address in the tutorial(s) and the exam board which you’re on and I’ll be sure to get back to you as soon as possible.
I look forward to hearing from you!
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|Further Mathematics||GCSE||£22 /hr|
|Philosophy & Ethics||A-Level||A|
Nadeem (Student) February 29 2016
Nadeem (Student) March 20 2016
Molly (Parent) February 29 2016
Pierre (Student) February 17 2016
This is a function which is in the form,
y = f(x)g(x)
It's the product of two functions and so we must make use of the product rule. This is a simple formula which you have to remember:
dy/dx = f'(x)g(x) + f(x)g'(x).
In words: the derivative of first function multiplied by the original second function, plus, the derivative of the second function multiplied by the original first function.
In this question,
f(x) = x
g(x) = sin(x)
so we can find that,
f'(x) = 1
g'(x) = cos(x)
and by substituting this into the formula for the product rule we get the answer:
dy/dx = sin(x) + xcos(x).see more