MYTUTOR SUBJECT ANSWERS

609 views

Understanding differentiation from first principle.

What is first principle? Why does it look nothing like the differentiation I had been doing for the past few years?

 

Differentiation from first principle is the main idea behind differentiation, a technique we employ to measure instantaneous rate of change. By now, you probably recognize "rate of change" as being synonymous to the term "gradient" or "average speed/distance" in context of several word problems. It is, therefore, not surprising that the definition of first principle is very similar to the gradient formula!

 

Let us recall: the gradient of line connecting two points (x1,y1) and (x2,y2) is given by the equation grad = (y1-y2)/(x1-x2). The definition of first principle says f'(x) = lim_{dx->0} (f(x)-f(x+dx))/dx. (Due to limitation of software, we'll refer to "delta x" by "dx" here for convenience.)

 

Let's examine my claim that both formulas are very similar. Suppose I have a graph y=f(x). Pick a point on the graph and name it (x1,y1). Now, to find the instantaneous rate of change at point (x1,y1), what can we do? Without the knowledge of the equation of the graph, we cannot perform differentiation using the rules that we'd learned. So, instead of trying to get the correct answer, let's approximate. Let's pick a point as cloes to (x1,y1) as we can and name this point (x2,y2). Because these points are very close together, it is safe to say that the line connecting them is a good approximation of the tangent line to point (x1,y1). In that case, we can use the gradient formula to find an approximate solution to the instantaneous gradient at (x1,y1)! By letting the distance between x1 and x2 be dx, we have: x2=x1+dx. Plug this into the gradient formula and we will get

approx. grad = (y1-y2)/(x1-x2) = (f(x1) - f(x2))/(x1 - (x1+dx)) = (f(x1) - f(x1+dx))/dx.

 

There is still something different - the "lim_{dx -> 0}" notation. What does it mean anyway? Reading the symbols out it says "the limit of the expression as "dx" approaches zero". You see, in order to improve accuracy of our gradient's approximation, we only need to pick (x2,y2) to be even closer to (x1,y1). Theoretically, then, if the distance between x1 and x2 is so close that it is almost zero, then our approximation would be the exact solution. Hence, if we allow "dx" to approach zero, then we can confidently change the left-hand-side of the approximation to the exact value, in other words:

f'(x1) = lim_{dx->0} (f(x1)-f(x1+dx))/dx. 

 

Since (x1,y1) is just a name we gave to the point that we are interested in, we may substitute it with (x,y) (to make the formula applicable in a generic x-y plot) to obtain the definition of first principle as presented in our textbooks. 

Jia Hao L. A Level Further Mathematics  tutor, A Level Maths tutor, G...

2 years ago

Answered by Jia Hao, an A Level Further Mathematics tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

62 SUBJECT SPECIALISTS

£20 /hr

Joel B.

Degree: Mathematics & Computer Science (Bachelors) - Manchester University

Subjects offered: Further Mathematics , Physics+ 3 more

Further Mathematics
Physics
Maths
Computing
Chemistry

“About Me: Hello, my name is Joel and I am currently a student at the University of Manchester studying for a joint honours in Mathematics & Computer Science. I have experience in teaching maths (and further maths) to students in the ag...”

MyTutor guarantee

£24 /hr

Barnaby W.

Degree: Mathematics (Masters) - Southampton University

Subjects offered: Further Mathematics , Maths

Further Mathematics
Maths

“ I have always enjoyed tutoring. I find that it gives me real satisfaction to see a child or student you have helped feel much more confident”

£22 /hr

Aldo E.

Degree: Engineering (Masters) - Cambridge University

Subjects offered: Further Mathematics , Physics+ 1 more

Further Mathematics
Physics
Maths

“Hi, I'm Aldo! I'm a Cambridge engineer and I'm passionate about maths and physics, as well as passing on knowledge to others!”

About the author

Jia Hao L.

Currently unavailable: for regular students

Degree: Pure Mathematics and Mathematical Logic (Masters) - Manchester University

Subjects offered: Further Mathematics , Maths

Further Mathematics
Maths

“There is nothing too difficult for you to understand with my help.”

You may also like...

Other A Level Further Mathematics questions

How do you calculate the derivative of cos inverse x?

How to determine the rank of a matrix?

Integrate cos(4x)sin(x)

Convert the general complex number z=x+iy to modulus-argument form.

View A Level Further Mathematics tutors

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok