I am a 3rd year Geophysics and Geology student at Durham University. Ever since I was small understanding the way things work, whether it be how mountains formed, the way planets move or solving a differential equation, has always been a passion for me. My aim is that, through my tutorials, you will also have the same love too. This interest and ability came in use when I was asked to be an academic mentor during my time at 6th form to help students lower in the school who needed additional help.
I have been a competitive swimmer from the age of 7 and used these skills to provide one to one swim coaching last summer for a 6 year old child. I also have over 200 hours sports coaching volunteering.
The content of each session is entirely guided by you and what you're needing help with at the time. I hope to make learning enjoyable and easy for you with a variety of different techniques (diagrams, exam questions, analogies) so that you can explain the concepts back to me just as well as I did for you!
Everybody learns differently and I will personalise the tutorials in order to help you learn to the best of your ability. I firmly believe that I can make each session useful and, more importantly, enjoyable! (even in the grueling exam time).
If you want to get into contact with me, send me a WebMail or book a Meet The Tutor Session. Just remember to tell me what subject you're studying and what exam board.
Cant wait for our first session!
|Geography||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Fatima (Student) October 5 2016
Using the laws of logs you can see that if you log both sides of the equation you get:
(2x+1)*log(8) = log(24)
Dividing both sides of the equation by log(8) you get:
2x+1 = log(24)/log(8)
Then it is a simple case of solving for x:
x = 0.5*(((log(24)/log(8))-1)
x = 0.264see more
Firstly, when approaching a differentiation question you need to work out what method you need to use to solve it. As you can see there are two terms multiplied by one another (the 'x' term and the '(x-2)-1/2' term), therefore the product rule must be used.
Making u = x and v = (x-2)-1/2
du/dx = 1 dv/dx = -1/2*(x-2)-3/2
Substituing these things into the Product Rule equation we get:
dy/dx = -x/2*(x-2)-3/2 + (x-2)-1/2
Now we need to focus on manipulating this equation to match the one given in the question. To start with we will take out a factor of (x-2)-3/2 giving:
dy/dx = (x-2)-3/2*(-x/2 + x-2)
dy/dx = (x-2)-3/2* (x/2 -2)
Multiplying by 2 :
dy/dx = (x-4)/2*(x-2)3/2see more