I am a Mathematics student at Warwick University. My enthusiasm for all things mathematical has no bounds and it's not just Maths that interests me but also the way that people learn it.
I am patient and friendly. Over the last few years, I've got involved with several programmes in which I helped teach a wide variety of students. From those who love Maths and want to do it for the rest of their lives to those who just want to get a good grade.
In our sessions, I will tailor everything we do to you. I will listen to what you want to get out of it and we'll proceed with that end goal in mind.
With Mathematics it's important that you understand the concepts you're being taught, it's then that you'll be able to use these ideas in exams - and in the future! With this in mind, I will go over these ideas thoroughly, helping you get to the point where you can explain these concepts to others with ease.
I hope the sessions will be enjoyable! The things you learn in the Maths A-level are amazingly interesting, hopefully if you don't see that now you will do by the end of a session.
If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session' - which you can do on this website. When you do so, tell me your exam board and what you want help with.
I look forward to meeting you!
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|STEP III||Uni Admissions Test||2|
|AEA||Uni Admissions Test||Distinction|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
This is a difficult question that you only need to know the result of.However, it's a good exercise to derive it.
Starting with f(x)=ax we can take the natural logarithm of both sides (so we can use one of its properties).
This gives us ln(f(x))=ln(ax), from the natural logarithms properties we know this is equal to ln(f(x))=x*ln(a).
Now using the chain rule we can differentiate both sides,
d(ln(f(x)))/dx= f'(x)/f(x), d(x*ln(a))/dx=ln(a)
so we now have f'(x)/f(x)=ln(a). Recalling that f(x)=ax this gives us the answer,
This is a typical further maths question, doing it correctly is a matter of carrying out a two-step process.
Start by finding the determinant of the matrix,
Then swap the entries a d and negate the other entries. After dividing by the determinant the inverse of A is given.
A^-1=1/(ad-bc)([d -b],[-c, a]).see more