**A bit about me:**

Hi! I'm Hannah and I have just received my Masters degree in Maths at Bristol University, graduating with a first. I have been a high achiever throughout my time in education and I now hope to help the next generation of students do the same, working in a** friendly, relaxed** atmosphere.

I had always gotten high grades in maths at school, but I had never really enjoyed it until I went to sixth form college. There I realised how **satisfying** maths can be and began to consider studying it at university. However at the time I wasn't taking further maths, so to catch up with the rest of my peers I** **learnt** **both maths and further maths modules** independently**, and was able to finish college with a further maths A level at grade A*. **I know the techniques needed **to understand these modules so am well equipped to explain them to you!

**The s****essions:**

**What we cover in sessions is up to you.** We will spend time on those areas of maths you find most challenging, and that are relevant to your specific exam board.

I believe that, particularly in maths, once the key concepts have been covered, the best way to improve your understanding is practice practice** practice**! Sometimes a topic doesn't click into place until you look at questions based on it. Therefore, during our tutor sessions I would aim to go through example questions in detail, until you feel **confident** enough to explain them to me.

To help you get the best out of your sessions I will make them as **engaging as possible.** Actively taking part in learning gives a greater appreciation for the subject (and achieves much **better results**!) than listening to someone ramble on.

**Any questions?...**

Feel free to contact me by sending a 'Webmail' or booking a 'Meet the tutor' session. We can discuss the areas you would like to work on and the best approach to your sessions, should you choose me as your tutor. I look forward to meeting you!

**A bit about me:**

Hi! I'm Hannah and I have just received my Masters degree in Maths at Bristol University, graduating with a first. I have been a high achiever throughout my time in education and I now hope to help the next generation of students do the same, working in a** friendly, relaxed** atmosphere.

I had always gotten high grades in maths at school, but I had never really enjoyed it until I went to sixth form college. There I realised how **satisfying** maths can be and began to consider studying it at university. However at the time I wasn't taking further maths, so to catch up with the rest of my peers I** **learnt** **both maths and further maths modules** independently**, and was able to finish college with a further maths A level at grade A*. **I know the techniques needed **to understand these modules so am well equipped to explain them to you!

**The s****essions:**

**What we cover in sessions is up to you.** We will spend time on those areas of maths you find most challenging, and that are relevant to your specific exam board.

I believe that, particularly in maths, once the key concepts have been covered, the best way to improve your understanding is practice practice** practice**! Sometimes a topic doesn't click into place until you look at questions based on it. Therefore, during our tutor sessions I would aim to go through example questions in detail, until you feel **confident** enough to explain them to me.

To help you get the best out of your sessions I will make them as **engaging as possible.** Actively taking part in learning gives a greater appreciation for the subject (and achieves much **better results**!) than listening to someone ramble on.

**Any questions?...**

Feel free to contact me by sending a 'Webmail' or booking a 'Meet the tutor' session. We can discuss the areas you would like to work on and the best approach to your sessions, should you choose me as your tutor. I look forward to meeting you!

No DBS Check

To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx)

To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x)

Firstly, using a standard result we have dy/du=1/u

Secondly, we must work out du/dx. Another standard result is that d/dx(sin(ax))=a*cos(ax) for any constant number a. This means du/dx=3*cos(3x)

Putting the two parts together, we find that our answer, given by dy/dx, is equal to

3*cos(3x)*1/u=(3*cos(3x))/(ln(sin(3x)))

To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx)

To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x)

Firstly, using a standard result we have dy/du=1/u

Secondly, we must work out du/dx. Another standard result is that d/dx(sin(ax))=a*cos(ax) for any constant number a. This means du/dx=3*cos(3x)

Putting the two parts together, we find that our answer, given by dy/dx, is equal to

3*cos(3x)*1/u=(3*cos(3x))/(ln(sin(3x)))

There are 2 methods we can use here.

**Method 1** - Factorisation

If we can rewrite the equation so that it looks like (x+a)*(x+b)=0, where 'a' and 'b' are numbers, then we can easily find a solution. If one or both of the terms inside the brackets equals 0, then their product is also 0. This means the solution(s) are the values of x such that x+a=0 or x+b=0.

Multiplying out the brackets of (x+a)*(x+b)=0 gives x^{2}+(a+b)*x+a*b=0. If we match up the terms with the equation we want to solve (called comparing coefficients), we see that we want two numbers 'a' and 'b' so that a+b=-5, and a*b=6. The two numbers which satisfy this are -2 and -3, so these are our values for 'a' and 'b'.

Now we can write our equation as (x-2)(x-3)=0 (you can check yourself this can be multiplied out to give the original equation!)

If x-3=0 then x=3, and if x-2=0 then x=2, so these are our two solutions.

**Method 2** - Quadratic Formula

This method can be used when you can't easily 'see' the numbers to use to factorise the equation.

The quadratic formula states that for an equation of the form a*x^{2}+b*x+c=0, where 'a', 'b', and 'c' are numbers, then

x=(-b+-sqrt(b^{2}-4*a*c)/(2*a)

For our equation a=1, b=-5 and c=6, which can then be substituted into the formula. Again you can check yourself that it simplifies to give the same answers as method 1

There are 2 methods we can use here.

**Method 1** - Factorisation

If we can rewrite the equation so that it looks like (x+a)*(x+b)=0, where 'a' and 'b' are numbers, then we can easily find a solution. If one or both of the terms inside the brackets equals 0, then their product is also 0. This means the solution(s) are the values of x such that x+a=0 or x+b=0.

Multiplying out the brackets of (x+a)*(x+b)=0 gives x^{2}+(a+b)*x+a*b=0. If we match up the terms with the equation we want to solve (called comparing coefficients), we see that we want two numbers 'a' and 'b' so that a+b=-5, and a*b=6. The two numbers which satisfy this are -2 and -3, so these are our values for 'a' and 'b'.

Now we can write our equation as (x-2)(x-3)=0 (you can check yourself this can be multiplied out to give the original equation!)

If x-3=0 then x=3, and if x-2=0 then x=2, so these are our two solutions.

**Method 2** - Quadratic Formula

This method can be used when you can't easily 'see' the numbers to use to factorise the equation.

The quadratic formula states that for an equation of the form a*x^{2}+b*x+c=0, where 'a', 'b', and 'c' are numbers, then

x=(-b+-sqrt(b^{2}-4*a*c)/(2*a)

For our equation a=1, b=-5 and c=6, which can then be substituted into the formula. Again you can check yourself that it simplifies to give the same answers as method 1