Hi, I'm Joe, an experienced tutor and a University of Bristol graduate with a first class Degree in Mathematics. Since joining mytutor, I've greatly enjoyed helping many students to gain a stronger understanding of mathematics and to excel. I try to make my tutorials engaging, and tailor them to your individual needs so that we are working towards the grade you aspire to achieve! I've had experience tutoring Maths at GCSE and A-level standard, working with students at different levels of understanding.

Hi, I'm Joe, an experienced tutor and a University of Bristol graduate with a first class Degree in Mathematics. Since joining mytutor, I've greatly enjoyed helping many students to gain a stronger understanding of mathematics and to excel. I try to make my tutorials engaging, and tailor them to your individual needs so that we are working towards the grade you aspire to achieve! I've had experience tutoring Maths at GCSE and A-level standard, working with students at different levels of understanding.

The main aim of these sessions is to provide you with the skills and confidence to overcome the parts of maths that you find difficult. Because of this you will choose the areas that we’re going to cover in each session and I will provide the necessary materials to facilitate. What Next? If you have any questions, please book a 'Meet the Tutor Session'! Then you can tell me your exam board, what modules you are studying (if you are an A-Level student) and what you find challenging. I look forward to hearing from you!

The main aim of these sessions is to provide you with the skills and confidence to overcome the parts of maths that you find difficult. Because of this you will choose the areas that we’re going to cover in each session and I will provide the necessary materials to facilitate. What Next? If you have any questions, please book a 'Meet the Tutor Session'! Then you can tell me your exam board, what modules you are studying (if you are an A-Level student) and what you find challenging. I look forward to hearing from you!

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Charlie (Student)

January 22 2018

great help, fantastic help with everything!

Charlie (Student)

December 6 2017

really good, really helpful!

Charlie (Student)

November 30 2017

Really good Really helpful

Charlie (Student)

August 31 2017

amazing help, helpful as always

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2).

Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)

By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).

When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2).

Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)

By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).

When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.

In order to deal with the modulus sign, we must take account of 2 possible cases:

Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1.

Case 2: |2x + 1| = -(2x +1). In this case we solve algebraicly again so that x < 4 + (2x +1) = 2x + 5. Hence by subtracting a 5 and an x from each side we get x > -5.

Finally we combine the results of each case, namely that x < 1 and x > -5 to get -5 < x < 1 as our final solution.

In order to deal with the modulus sign, we must take account of 2 possible cases:

Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1.

Case 2: |2x + 1| = -(2x +1). In this case we solve algebraicly again so that x < 4 + (2x +1) = 2x + 5. Hence by subtracting a 5 and an x from each side we get x > -5.

Finally we combine the results of each case, namely that x < 1 and x > -5 to get -5 < x < 1 as our final solution.