Hello! I'm a first year Computer Science student at the University of York.
I offer tutoring for maths at GCSE and A-Level. Despite not studying maths now, it was my original love - the subject which first taught me how to enjoy solving the even the most devious of problems. Through my tutoring sessions, I hope to inspire some of the same enthusiasm for maths in my students.
The tutoring sessions are quite flexible and can be changed for the tutee's needs. I am happy to work through a problem with them that they are stuck on. Alternatively, the tutee can suggest a topic they want practice in, and I can work through common questions with them.
Feel free to send me a message! Tell me what topic/question you're struggling with, and it would also be helpful if you tell me what exam board you study.
|Maths||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Barbara (Parent) January 25 2017
Abbie Hollis (Student) January 13 2017
Abbie Hollis (Student) January 5 2017
Barbara (Parent) December 2 2016
In order to differentiate this expression, you need to use the chain rule.
The chain rule gives: f'(x) = uv' + u'v.
The u and the v are two parts of the original function f(x). The apostrophe ' at the end means the derivative of that.
We need to assign values to u and v, so we look at the function f(x) = 2x(3x - 1)2 to see what parts it is in:
u = 2x
v = (3x - 1)2
Then, we differentiate each of these.
u' = 2
v' = 2 x 3 x (3x - 1)1 = 6(3x - 1)
Now, we can put this expression altogether:
f'(x) = uv' + u'v = 2x(6(3x - 1)) + 2(3x - 1)2
And now, simplify.
f'(x) = 12x(3x - 1) + 2(3x - 1)2
f'(x) = 2(3x - 1)[6x + (3x - 1)]
f'(x) = 2(3x - 1)(9x - 1)
f'(x) = 2(27x2 - 9x - 3x + 1)
f'(x) = 2(27x2 - 12x + 1)
f'(x) = 54x2 - 24x + 2see more