John H.

Currently unavailable: for regular students

Ship Science (Masters) - Southampton University

4.0

Trusted by schools

5 completed lessons

Who am I?

A Ship Science (Engineering!) student at the University of Southampton, I have always loved science, maths and computers. I enjoy sharing that love, so I will do my best to show you why I think these things are so exciting!

I'm a pretty experienced tutor, having taught everything from sailing and navigation through to first aid before, and I believe that the key to effective learning is fun. You will never learn if you're bored and hate your subject, so I will keep the sessions varied and interesting to make sure you're always excited by what you learn!

Maths and the Sciences are the basis of everything we see and do, and understanding them allows us to understand and manipulate our universe in amazing ways.

Understanding is the key to all science and maths questions, so when you come to me with a question you're struggling with I'll first help you take the question apart and see the basic concepts that apply to it, then walk you through piecing them back together to complete the question, and any others like it.

Sound good?

If you think I can help, send me a message on here or book a free Meet the Tutor session to discuss what you're struggling with and how we can fix it together.

Can't wait to meet you!

Who am I?

A Ship Science (Engineering!) student at the University of Southampton, I have always loved science, maths and computers. I enjoy sharing that love, so I will do my best to show you why I think these things are so exciting!

I'm a pretty experienced tutor, having taught everything from sailing and navigation through to first aid before, and I believe that the key to effective learning is fun. You will never learn if you're bored and hate your subject, so I will keep the sessions varied and interesting to make sure you're always excited by what you learn!

Maths and the Sciences are the basis of everything we see and do, and understanding them allows us to understand and manipulate our universe in amazing ways.

Understanding is the key to all science and maths questions, so when you come to me with a question you're struggling with I'll first help you take the question apart and see the basic concepts that apply to it, then walk you through piecing them back together to complete the question, and any others like it.

Sound good?

If you think I can help, send me a message on here or book a free Meet the Tutor session to discuss what you're struggling with and how we can fix it together.

Can't wait to meet you!

Enhanced DBS Check

31/03/2015

#### Ratings & Reviews

4from 1 customer review

Amy (Student)

December 5 2016

Really good except the internet was a bit dodgy

#### Qualifications

MathsA-level (A2)A*
Further MathsA-level (A2)B
Physics A-level (A2)B
ComputingA-level (A2)A

#### General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

#### Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
ComputingGCSE£18 /hr
ICTGCSE£18 /hr
MathsGCSE£18 /hr
MathsIB£20 /hr
Maths13 Plus£18 /hr

### What exactly IS differentiation?

This is a common question as we are often taught to differentiate by simply being told how to do the calculation, but not what it is we're really doing.

The concept

Differentiation is finding the gradient of a line. For simple problems this is easy to see using the normal trignometry methods for finding a gradient: Draw a graph of y = x and you can see that the gradient of the line is 1. Integrate y = x and you find that dy/dx = 1, as you would hope!

However the trignometric methods will not work for a curve such as y = x2 as the gradient is different at every point, so we have to use the differentiation. In this case what the differentiation is really doing is finding the gradient of the line at any given point x. This can be understood through differentiating by first principles.

First Principles

To differentiate y = x2 from first principles we begin by finding the gradient between point (x, x2) and an arbitrary point (x+h, (x+h)2).
This gives us the following equation:

$\frac{dy}{dx}&space;=&space;\frac{x^{2}-(x+h)^{2}}{x-(x+h))}$
which simplifies to:

$\frac{dy}{dx}&space;=&space;2x+h$

Now for the important bit. We found the gradient between x and another point a distance h from x, but we want the gradient at x, so we find the limit of the above equation when h tends to 0. You can see on the graph below how reducing the value of h to 0 will give us the gradient at point x:

In our case of y = x2, this gives us the following answer:

$\inline&space;\frac{dy}{dx}&space;=&space;\lim_{h\rightarrow&space;0}(2x&space;+&space;h)=2x$

Which gives us exactly what we would get by the differentiation methods we know!  You can try this with almost any differential function and you'll find that it works, but don't try anything too complicated just yet, as sometimes it can be tricky to evaluate that limit!

This is a common question as we are often taught to differentiate by simply being told how to do the calculation, but not what it is we're really doing.

The concept

Differentiation is finding the gradient of a line. For simple problems this is easy to see using the normal trignometry methods for finding a gradient: Draw a graph of y = x and you can see that the gradient of the line is 1. Integrate y = x and you find that dy/dx = 1, as you would hope!

However the trignometric methods will not work for a curve such as y = x2 as the gradient is different at every point, so we have to use the differentiation. In this case what the differentiation is really doing is finding the gradient of the line at any given point x. This can be understood through differentiating by first principles.

First Principles

To differentiate y = x2 from first principles we begin by finding the gradient between point (x, x2) and an arbitrary point (x+h, (x+h)2).
This gives us the following equation:

$\frac{dy}{dx}&space;=&space;\frac{x^{2}-(x+h)^{2}}{x-(x+h))}$
which simplifies to:

$\frac{dy}{dx}&space;=&space;2x+h$

Now for the important bit. We found the gradient between x and another point a distance h from x, but we want the gradient at x, so we find the limit of the above equation when h tends to 0. You can see on the graph below how reducing the value of h to 0 will give us the gradient at point x:

In our case of y = x2, this gives us the following answer:

$\inline&space;\frac{dy}{dx}&space;=&space;\lim_{h\rightarrow&space;0}(2x&space;+&space;h)=2x$

Which gives us exactly what we would get by the differentiation methods we know!  You can try this with almost any differential function and you'll find that it works, but don't try anything too complicated just yet, as sometimes it can be tricky to evaluate that limit!

2 years ago

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