Degree: Mathematics with Mathematical Physics (Bachelors) - University College London University
I am an undergraduate student at UCL studying Mathematics with Mathematical Physics. Hoping to go into research after my degree, I have a true passion for my subjects and my aim is to pass on that passion to students in a friendly, patient manner. I also have a love for playing piano and practicing martial arts, both of which I have been doing regularly for a couple of years.
My Approach To Teaching Maths:
Mathematics nowadays may seem to some students like 'that subject for which they have to memorize all these formulas for'.
Speaking from personal experience, I can say that mathematics can and should be taught as a subject that relies almost entirely on understanding and seldomly on memorization. The reasons for this are numerous, but here are two of the reasons I find most important:
Firsly, mathematics truly can be a beautiful subject which not only has many, many practical applications in other areas, but also as a standalone subject produces many beautiful results.
Secondly, the frightening amount of 'rules' that one would have to memorize to do well in school can reduce to only a handful of basic results, from which all else can be derived, by having a good understanding of the basic concepts.
I am therefore very keen to be able to pass on some of my knowledge to both give students a better understanding of the mathematics they use and naturally, make their life easier by allowing them to exchange memorization by comprehension.
I hope that I have convinced you that this way of teaching is the one which will prove to be most fruitful and I look forward to meeting you!
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Let us denote sin(nx) = u(x), where u is a function of x. The equation is now therefore f(x) =(u(x))^n.
For simplicity, we will write that as f(x) = u^n
By the chain rule, we know that f'(x) = df/dx = (df/du)*(du/dx).
Firstly computing df/du, we find df/du = n*u^(n-1)
Now we need to find du/dx. Since u = sin(nx) , du/dx = ncos(nx).
Therefore, our answer is f'(x) = (df/du)*(du/dx) = n*u^(n-1)*ncos(nx),
subbing in u = sin(nx) yields the final answer:
f'(x) = n(sin(nx))^(n-1)*ncos(nx)see more