__About Me:__

Hi! My name's Ethan, and I'm currently a first year Engineering Maths student at Bristol University. I've always had an ambition to be able to solve problems, which is probably why I'm keen on maths!

I'm a really patient person, and I understand that maths doesn't come easily to everyone! I'm also a certified instructor in Tae-Kwon-do, teaching children from as young as four up to adults, so I have plenty of experience in driving people on to what they want to achieve.

__During the Sessions:__

During a session, you'll be in charge of what we cover. The key part of maths is being able to understand the basic concepts and then build on them.

I'm always thinking of ways to make maths more fun myself, and hope I can pass it on through the sessions! I know that some methods don't work for everybody, so I'll always have something else prepared and ready to use, so by the end of it, you should be able to explain it back to me!

At the end of the day, my aim is to make it enjoyable as well as to get you to where you want to be. Maths has a countless number of uses, and it’s an incredible topic to learn, so I hope I can get my enthusiasm across to you!

__What now?__

If you have any kind of questions, feel free to send me a message or book a 'Meet the Tutor' session, both accessible through the website.

Don't forget to let me know what you're struggling with, and which exam board you're on.

I look forward to meeting you!

__About Me:__

Hi! My name's Ethan, and I'm currently a first year Engineering Maths student at Bristol University. I've always had an ambition to be able to solve problems, which is probably why I'm keen on maths!

I'm a really patient person, and I understand that maths doesn't come easily to everyone! I'm also a certified instructor in Tae-Kwon-do, teaching children from as young as four up to adults, so I have plenty of experience in driving people on to what they want to achieve.

__During the Sessions:__

During a session, you'll be in charge of what we cover. The key part of maths is being able to understand the basic concepts and then build on them.

I'm always thinking of ways to make maths more fun myself, and hope I can pass it on through the sessions! I know that some methods don't work for everybody, so I'll always have something else prepared and ready to use, so by the end of it, you should be able to explain it back to me!

At the end of the day, my aim is to make it enjoyable as well as to get you to where you want to be. Maths has a countless number of uses, and it’s an incredible topic to learn, so I hope I can get my enthusiasm across to you!

__What now?__

If you have any kind of questions, feel free to send me a message or book a 'Meet the Tutor' session, both accessible through the website.

Don't forget to let me know what you're struggling with, and which exam board you're on.

I look forward to meeting you!

Enhanced DBS Check

28/05/20124.5from 2 customer reviews

Dawn (Parent)

February 15 2015

It was a very helpful session explained the subject well.

Paula (Parent)

February 9 2015

Great informative lesson, much happier going into test the next day

The chain rule is used when you're trying to differentiate a complicated function, and it allows you to split up the function into easier ones. The trick is spotting where to split the function, and what part you should substitute.

Say, for example, you're trying to find the derivative of: y=(1+x)^2 . If you can replace a suitable function of x with a function of u, you would be left with y=f(u) and u=g(x), y and u being functions of u and x respectively.

Now, if you differentiate y with respect to u, and u with repsect to x, you're left with the equation: dy/dx = dy/du * du/dx . On the RHS, you're left with dy/dx, as the du's cancel out, which is what you were after at the beginning. All that you would have to do is replace any remaining u with your function of x to find your desired derivative.

It's better understood through an example. Let's have a look at the example above, y=(1+x)^2. Take u=1+x, and youre left with y=u^2. Now, differentiate both fuctions, y=f(u) and u=g(x), and you're left with: dy/du=2u and du/dx=1.

Now, from the formula above, we know that dy/dx=dy/du * du/dx. Therefore: dy/dx = 2u * 1 = 2u. Now all that's left to do is replace u with the function of x you originally took out. In this case, u=1+x, so: dy/dx = 2(1+x) .

For a slightly harder example, try: y=1/(x^2+4x)^1/2 . Rearrange this to have y in terms of powers, similar to the first example: y=(x^2+4x)^-1/2 .

Now, take an appropriate function of x to equal u. In this example, it would be a good choice to take u=x^2+4x , leaving y=u^-1/2. Differentiating both of these functions gives: dy/du=-1/2(u)^-3/2 and du/dx=2x+4.

We know that dy/dx=dy/du * du/dx, therefore: dy/dx = -1/2(u)^-3/2 * (2x+4) . Replacing u with the function of x you took out gives:

dy/dx = -1/2(x^2+4x)^-3/2 * (2x+4) .

Rearranging and simplifying this expression gives:

dy/dx = -(x+2) / (x^2+4x)^3/2 .

The chain rule is used when you're trying to differentiate a complicated function, and it allows you to split up the function into easier ones. The trick is spotting where to split the function, and what part you should substitute.

Say, for example, you're trying to find the derivative of: y=(1+x)^2 . If you can replace a suitable function of x with a function of u, you would be left with y=f(u) and u=g(x), y and u being functions of u and x respectively.

Now, if you differentiate y with respect to u, and u with repsect to x, you're left with the equation: dy/dx = dy/du * du/dx . On the RHS, you're left with dy/dx, as the du's cancel out, which is what you were after at the beginning. All that you would have to do is replace any remaining u with your function of x to find your desired derivative.

It's better understood through an example. Let's have a look at the example above, y=(1+x)^2. Take u=1+x, and youre left with y=u^2. Now, differentiate both fuctions, y=f(u) and u=g(x), and you're left with: dy/du=2u and du/dx=1.

Now, from the formula above, we know that dy/dx=dy/du * du/dx. Therefore: dy/dx = 2u * 1 = 2u. Now all that's left to do is replace u with the function of x you originally took out. In this case, u=1+x, so: dy/dx = 2(1+x) .

For a slightly harder example, try: y=1/(x^2+4x)^1/2 . Rearrange this to have y in terms of powers, similar to the first example: y=(x^2+4x)^-1/2 .

Now, take an appropriate function of x to equal u. In this example, it would be a good choice to take u=x^2+4x , leaving y=u^-1/2. Differentiating both of these functions gives: dy/du=-1/2(u)^-3/2 and du/dx=2x+4.

We know that dy/dx=dy/du * du/dx, therefore: dy/dx = -1/2(u)^-3/2 * (2x+4) . Replacing u with the function of x you took out gives:

dy/dx = -1/2(x^2+4x)^-3/2 * (2x+4) .

Rearranging and simplifying this expression gives:

dy/dx = -(x+2) / (x^2+4x)^3/2 .