Sam E.

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Degree: Computer Science (Masters) - York University

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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.

Tutoring sessions:

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.

Contact me:

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.

#### Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr

#### Qualifications

MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ComputingA-LevelA*
 CRB/DBS Standard No CRB/DBS Enhanced No

#### General Availability

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#### Ratings and reviews

4.8from 4 customer reviews

Barbara (Parent) January 25 2017

Abbie Hollis (Student) January 13 2017

Abbie Hollis (Student) January 5 2017

Barbara (Parent) December 2 2016

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### How would you differentiate f(x) = 2x(3x - 1)^2 using the chain rule?

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 fu...

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 + 2

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7 months ago

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