Currently unavailable: for regular students
Degree: Mathematics, Operational Research, Statistics and Economics (Bachelors) - Warwick University
I study MORSE at Warwick University (unfortunately no, it's not morse code) and hope that my expertise in both teaching and learning maths (as well as a couple of other things) will translate into a great learning experience for you.
I've been tutoring for a couple of years and have also directed theatre productions, so I have a fair bit of experience with direction and teaching.
In these I'd hope that you provide me with an area/section (don't worry if it's the whole module!) you wish to cover, then we'd work through a foundation of understanding, and I will use as many analogies or sketches or what have you to try and best explain my thought process and help you understand the content. After a solid foundation is built, we'd move on to tackle exam and exam-style questions - and perhaps even slightly beyond that, to really affirm your knowledge.
I look forward to working with you!
|Further Mathematics||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|English Literature||GCSE||£20 /hr|
|English||13 Plus||£20 /hr|
|Maths||13 Plus||£20 /hr|
|English||11 Plus||£20 /hr|
|Maths||11 Plus||£20 /hr|
|-Personal Statements-||Mentoring||£22 /hr|
|.MAT.||Uni Admissions Test||£25 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Beverley (Parent) April 10 2016
Hussain (Student) April 11 2016
Safwan (Student) April 27 2016
Hussain (Student) April 29 2016
The first thing to recognise is that this function is a product of two functions: namely, 4x^2 and ln(x), thus we must employ the product rule in order to find the solution. As you may recall, the product rule states that when you have a function f(x) = uv, the differential f'(x) = udv + vdu, thus:
we differentiate once, finding that dy/dx = (4x^2)/x + 8xln(x) and simplify to get the expression 4x + 8xln(x)
then differentiate a second time, remembering to once again employ the product rule for the second term in the equation:
d^2y/dx^2 = 4 + (8 + 8ln(x))
now substitute the value of x = e^2 into the equation:
thus d^2y/dx^2 = 12 + 8ln(e^2)
now as we know that the natural logarithm "ln" is the inverse of the exponential function "e", this becomes:
d^2y/dx^2 = 12 + 8(2)
= 28.see more