Differentiation: Introduction


Uploaded by numericalmethodsguy on 26.01.2009

Transcript:
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In this segment, I’m going to just have a brief differentiation
primer for you.
There’s a short primer available on the website,
and you can see that in the slate which was shown
in the video.|So going back to differentiation, in order
to be able to understand numerical differentiation, we do need to talk about what you learned in differential calculus class.
You already know that if you are trying to find out what the derivative of a function, f, is,
that is defined as limit, delta x approaching 0,
f of x plus delta x minus f of x
divided by delta x.|So that was the definition of your
derivative of a function, which from a graphical standpoint of view means
that if you had a function like this, so you have f of x
as a function of x, and you want to find out what the derivative
of the function here is, let’s suppose, or let’s suppose you’re trying to find the derivative of the function right here.
So this is x, let’s suppose.|You want to find the derivative of the function.
So what you’re going to do is, if you look at from the definition which you have here, that you’re going to choose a
point which is delta x ahead of this point, so the distance between
the two points right here is delta x.|And what you’re going to do is you’re going to
draw the line between the value of the function at
x and x plus delta x, that’s what you’re doing here, and then you’re going to take this
line, which is the other line right here.|Now if you look at the distance which is here
between those two, so this is the value of the function at x, this is the
value of the function at f of x plus delta x, so that distance becomes f of x plus
delta x minus f of x, and this distance which you are seeing
here is just delta x right there.|So we can
see that the slope of this particular line here . . . the slope
of this particular line here is this rise, that’s the rise,
and that’s the run.|So that’s what you’re seeing, the rise here and the run here,
but to define the value of the derivative exactly, what you want to do is you want to make
this delta x approaching 0.|And as we know in numerical methods, you cannot have . . .
you don’t have the luxury of choosing delta x approaching 0, you can only make delta x
to be a finite number, but you do need to understand what the exact definition of
the derivative of a function is.|And that’s what it is by making delta x approach
0, you’re going to get the numerator divided by denominator to be the derivative of the function
there.|Let’s take an example, you must have seen, let’s suppose if you have the function
f of x equal to 5 x squared.|From a differential calculus
class you already know f prime of x is what?|5 times the derivative of
x squared, which is 2 x, and I get 10 x.|And then
if you want to calculate the value of the derivative of the function at any point, any
point x, you can always plug in the value of x in there.|So for example if I’m trying to find out the value of
the function at 6 let’s suppose, I get 10 times 6, which is
60.|So that’s how we are able to calculate the derivative of a function at a particular
point by using your differential calculus knowledge, and then simply substituting the value of x
into the formula.|Now this 10 x which you are finding out here does not magically
appear there, but these are formulas which have been derived based on the
basic fundamental definition of the derivative of the function, which is right here.
Let me go back through that definition which is right here,
which is that you have the rise, the run, and you are
choosing delta x approaching 0.|So what I’m going to do is, I’m going to use this definition itself to
show you how do we get 10 x, as opposed to simply using the formula which I have learned in the differential
calculus class.|So let’s suppose if I had f of x which I have somebody
gave me to be 5 x squared, so I’m going to say f prime of x will be equal
to limit of delta x approaching 0,
f of x plus delta x minus f of x divided by delta x,
that’s’ the definition of f prime of x, so I’m just rewriting it again.
So for this particular function I have limit of delta x approaching 0, and what is
f of x plus delta x?|It will be 5 x plus delta x, whole squared.|Because
all it is, is what is the value of the function 5 x squared at the point x plus delta x.
Now what’s the value of the function at x?|It’s just 5 x squared, divided by delta x.
So I go and try to
expand this, I’ll use the a plus b whole squared formula, I’ll get x squared
plus delta x whole squared plus 2 x delta x
minus 5 x squared, divided by delta x.
So this expansion which you are seeing here is simply the formula a plus b whole squared is a squared
plus b squared plus 2 a b.|So you can very well see that what is happening is that
this is cancelling with this, because that’s 5 x squared, so you’re left with limit
of delta x approaching 0, 5 times delta x whole squared
plus 10 times x times delta x, divided by delta x,
that’s what you’re getting there.|If I’m going to divide
the numerator as well as denominator by delta x, I’m going to get limit of delta x
approaching 0, I’ll get 5 times delta x plus 10 times x,
because here, one delta x will be gone, and here, this delta x will
cancel with this delta x.|And now if I take the limit, the limit of 5 times
delta x will be just 0, and the limit of 10 x as
delta x approaches 0 will be 10 times x, and I’ll get 10 times x.
So that’s what all these derivatives are coming from, so when you use
the formulas from your differential calculus class to calculate the exact derivatives,
those have been calculated by using the fundamental definition of
the derivative of a function.|And that’s the end of this
segment.
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