BAM!! Mr. Tarrou In this video we are going
to learn something extremely cool. It a major
branch of calculus. It is called the derivative.
And what the derivative is going to do in
Calculus along with the tangent line problem
is we are going to be able to combine a very
old idea that you have been using, and learning,
and working with since Pre-algebra. But if
you combine that old slope of rise over runs
with our idea of limits that we just started
going through in the first chapter something
really cool happens. It allows you to find
the slope of a curve at any given point (almost).
It is called the slope of the tangent line.
Now in this video, this is going to be a full
length... many many... about a 9 or 10 example
video of anything hopefully you will need
to get through your first section of learning
how to the derivative with the definition
of derivative and the tangent line problem.
We are going to be finding the slope of the
tangent line at a point. We are going to be
finding the derivative function, the equation
that will allow us to find the slope at any
given point along the curve. That kind of
sounds like I said the same thing, but when
you use the definition of the derivative to
find the slope of tangent at a point and you
want the slope somewhere else in the curve
you have to go through the entire process
again. Whereas if you just find the derivative
function, that function will allow you to
put an pretty much... almost any x value and
find the slope at that point. But the definition
of derivatives starts to have some issues
when you have a derivative which is undefined.
So we are also going to be taking a look at
examples of finding the derivative of a vertical
line, trying to find the derivative at a sharp
bend. We are going to be talking about the
relationship of finding the derivative of
a function and continuity at a particular
point. There is a relationship there and we
will take a look at that as we try and find
the derivative of a Floor function or Step
function, the Greatest Integer function. We
are also going to be working through some
examples near the end of the video where I
help you find the tangent line to a curve
that goes through another point not on the
curve. And a tangent line which is parallel
to a given tangent line. So we have got a
lot to talk about. But the first thing I want
to do is turn off the air conditioner so yo
udon't here that in the back ground and show
you how the limit process will allow us to
work with two points on a curve and find an
approximation of the slope of the tangent
line. It will be pretty darn accurate with
our idea of limits. Alright. So what we are
looking here is a curve with two points and
I am going to draw a line through these two
points and 
the slope of this secant line, because it
is intersecting at more than one point, the
slope of that secant line... in precalculus
we would call the average rate of change.
It is supposed to be there to estimate the
direction or the slope of this curve at that
point. I want to work with finding the slope
at that point, not necessarily at this point
with the much more complicated coordinates
of c plus delta x f(c+Dx). That line, how
well do we think this line estimates the slope
of this curve at that point? It seems like
the line is steeper. That slope would be a
greater number than... it kind of looks like
the curve is going off in this direction.
So that line is an estimate of the slope of
this curve at that point, but it is not a
really good estimate. So what we are going
to do is, with again this idea of limits,
and I will summarize this up here with some
nice notes in a second that I will reveal
to you. I want to take this delta x and reduce
it. Like if this is a estimation of the slope
of this curve at this point, I can get a better
estimate of I move that point closer. So maybe
instead of having my point here, I will reduce
my delta x... this horizontal distance by
a little bit... and maybe we will put our
point here. Well if I redraw that secant line
the best I can without a ruler. We can see
that this slope of that line is starting to
come down a bit and do a better job of estimating
the slope of that curve at that point. But
it still seems like... A tangent line is ultimately
what I am trying to find, the slope of a tangent
line. A tangent line is only supposed to touch
the curve at one point. But you still need
two points to find the slope of a line. But,
those points are not really close enough yet
to give me a really good estimation. So what
we are going to do is, well I am going to
move my point even closer. Reduce that delta
x, i didn't really show it there, this is
my delta x now compared to my original one.
It is smaller. If I reduce that delta x to
an even smaller amount and move the point
even closer. Well now I have a secant line
because I am still identifying two points.
But now I have a secant line that has a slope
that is much closer to what appears to be
the slope of the curve at that point (c,f(c)).
Now what I am going to step off here and reveal
to you, is some notes about how we can...
remind us of what our old slope formula looks
like, move it into function notation, and
then introduce these variables here and show
you how limits can allow that point... that
delta x... I am going let that delta x reduce
to zero. Remember doing just that last chapter,
numerical and graphical limits and then we
learned about the properties of limits to
set these things up to allow this delta x
to approach some value. Well we are going
to allow that value to approach zero and get
those points infinitely close and basically
get a tangent line to the curve. It is really
technically a secant line because we are using
two points. But the distance between is going
to approach zero, and thus it is basically
a tangent. nanananana... Reduce delta x until
the points are infinitely close and you have
a tangent line instead of a secant line. The
slope of the tangent line is the slope of
the curve at that point. Now let's just build
up to the formula for a minute instead of
giving it straight to you. We have our old
school slope formula Y2-Y1 over X2-X1. In
more fancy pants type language that is delta
y over delta x. The change in y over the change
in x. Or as we have always known it rise over
run. If you remember from Precalculus or maybe
your Algebra 2 Honors class you might have
seen this formula for slope written as f(X2)-f(X1)
replacing the y sub 2 minus y sub 1. That
was just to get you used to using the function
notation, or try to make you comfortable with
using function notation. Also if you are again
near the end of your Algebra 2 honors class
or PreCalculus, you might of had this formula
call the average rate of change. It is the
slope of the secant line. it is the average
rate of change between two points on a curve.
We had to call it average rate of change because
the slope of a curve is always changing. Well
now again do I have these specifically labels
of (c,f(c)) and (c+Dx,f(c+Dx)) Well, this
point is fixed. We are going to basically
write a formula, it is called the definition
for the slope of the tangent line, and I want
to find the slope of this curve at that point.
So I want this point to be fixed and I want
this other point to have some identification
in it that relates it to this fixed point.
I can then reduce that delta x as I said,
reduce it to zero moving that point closer
to this one getting a more and more and more
accurate estimate of the slope. The slope
of the tangent line. So I am going to replace
that generic X2 with x+Dx and the X1 with
just c. Well now my denominator instead of
being X2-X1 it is c+dx-c. The c minus c cancels
out and we get delta x... Dx. Sorry for the
interruption there. So this formula was given
in your book not just as a random formula
that you used to practice your function notation,
but as a precursor to what we are about to
learn which is the definition of the slope
of the tangent line. We are going to take
this old school slope, we are going to introduce
limits, and BOOM! There is our first main
concept of this video. BAM! Here it is. The
main idea behind the entire video and the
many many examples yet to follow. Definition
of Tangent Line with Slope m. If function
f is defined on an open interval containing
c, and if the limit exists... this limit here...
the limit as delta x approaches zero of delta
y over delta x. Or more formally with the
difference quotient from PreCalculus, the
limit as delta x approaches 0 of f(c+Dx)-f(c)
over Dx. That is going to be the slope of
the tangent line. Well the line passing through
(c,f(c)), we are going to call that the point
of tangency, with slope 'm'... or I might
put a little subscript of tan... is the tangent
line to the graph of f at the point of that
tangency. Now it does say if this limit exists.
We just talked about limits in the previous
chapter. We know that maybe this ends up being
undefined and just goes to infinity. That
is going to end up being, or identifying the
slope of a vertical line. We will be doing
examples of that later. The left and right
limits may not be equal. Whether there is
a sharp bend or maybe the graph is just got
a non-removable discontinuity at this value
of c which we are finding the slope at. Now
these first three examples, again I have posted
them in an earlier video. I think it is called
Slope of a Tangent Line Derivative at a Point.
I am going to walk through the first example
step by step. Then because you can find these
two other examples explained step by step
by step, I will just reveal them to you with
a pause if you would like to do those problems
on your own. I will just reveal the solution
all at once. We have f(x) is equal to 3x+5.
I want to find the slope of this function
at the point of tangency of (2,11) Well c
is that value of x, that value along the x
axis at which we are trying to find the slope
of the tangent line. So when I say that I
want to find the slope of that tangent line
at the point of tangency of (2,11) I am telling
you that c is 2. So we are going to say that
the slope of the tangent line is equal to
the limit as delta x approaches 0 of... Now...
ok.. it is f(c+Dx) so here is my function
and we are going to do 3 times x which is
going to be again c+Dx, so it is going to
be 2+Dx. Now plus 5... minus... These parenthesis
I am putting around the first term in my numerator
honestly are a little bit unnecessary. But
because we have a minus sign here you better
make sure that if you have more than one term
in your function... you are pretty much guaranteed
to have that be the case... make sure you
wrap that second term in parenthesis to remember
to multiply that negative sign through the
parenthesis and not make a sign error. So
minus... Now let's see here. The formula says
f(c), so it is going to be this function with
2 plugged in for x.. all over delta x. Now
if I try and... We talked about in a previous
chapter finding limits both numerically and
graphically, and then we got tired of making
all of those tables and graphs. So we learned
some properties of limits. We used properties
of limits to find limits with direct substitution
when possible. Well I can't take this value
of zero and plug it into my expression because
this is going to be undefined. And... uh...
So when you do the definition of the derivative,
a lot of times... unless the limit does not
exist because of maybe a vertical tangent
line... that denominator will cancel out after
a little bit of simplification and manipulation
of this expression until you can take this
zero and plug it into the expression. That
is going to look like this. The limit as delta
x approaches zero of... let's see here. We
distribute this 3 and we get 6 plus 3 delta
x plus five. And then over here these are
just numerical values so we can work this
out. 3 times 2 is 6, now plus 5 is 11. Now
that all simplified to just be one term. So
I don't need to worry about distributing that
negative sign. But a lot of times you will,
especially when we get into finding the derivative
function as opposed to just a numerical slope
at a particular point. So that is going to
be again minus 11. All this over delta x.
Well, 6+5 is 11, and minus 11 so all of those
constants are going to cancel out. So again
6 plus 5 is 11, minus 11 is 0. We are just
left with the limit as delta x approaches
0 of 3Dx/Dx. These delta x's cancel out and
we end up with the limit as Dx approaches
zero is just a constant of 3. We know from
one of our properties of finding limits that
the limit of a constant is going to always
be equal to 0... Excuse me, not zero but 3.
I had a little brain fart there for a second.
No matter what Dx is approaching the function's
value is 3. It is 3. It is always equal to
3. That makes sense because look at what our
original function is. It is a line, y=mx+b.
This is a line. It is a linear function. The
slope is the same everywhere no matter what
the value of x is. No matter where you go
along that line you are going to have a slope
equal to 3. So this question showing the process
of working through the definition of the slope
of the tangent line. It is obvious that it
works because we got three and we can see
that slope in the original function. But this
original function is not a curve and the cool
thing about this definition of tangent line
with slope 'm', I am going to just call this
the definition of a derivative, is you can
find the slope of a curve at any point (almost).
So let's get this erased. I will put up the
previous question, and let's get to it right
now. For our next example we have 2x^2-3x+1.
We are going to find the slope of this parabola
at (2,3) This question is going to work just
like the last one, but you will have a little
bit of algebra work that is necessary because
we have 3 terms and especially since one of
them is squared. So I will get us started
and then I will walk off, give you about 5
seconds if you would like to pause the video
and finish it on your own before revealing
the answer. If I feel like there are any important
points I will point them out, and then get
on to the next example. So we want to find
again the slope of the tangent line at this
point. It is going to be the slope of the
tangent line is equal to the limit as delta
x approaches 0. Again I want to take this
function and plug in 'c' which is the x coordinate
at the point of tangency where we want to
find the slope at plus delta x. So it is going
to be 2, I don't need those parenthesis there.
2 times 'c' which is 2+Dx squared minus x
times 2+Dx, not squared this time cuz it is
just x... plus 1... minus.. Now I have better,
with these three terms and... Again I am always
going to be in the practice of using the minus
with parenthesis. But, we are going to plug
in a constant value of 2 so this will all
work out to a single number like 4. We have
got 2 times 2 squared minus 3 times 2 plus
1. All of this over again delta x. Now these
are pretty long questions when you use the
definition of a derivative to find the slope
of the tangent line at a point or to find
the actual derivative function itself. They
are long and tedious problems with a lot of
algebra. It is a really good review of all
your algebra skills that hopefully you remember
from last year. If not, you have a good opportunity
to relearn them. But they are long and tedious.
And there are plenty of places to make some
small mistakes. If you show your work. If
you just do one algebraic step in every line,
and you keep writing step after step after
step. Don't try to do too much in your head.
You will make fewer mistakes that way. And
if you do make a mistake, like a sign error
or something like that, it will be easier
to find and easier to fix. A lot less time
consuming. So don't shortcut on showing your
work for these problems. It will help you
tremendously to get these right more consistently.
I am going to step off for a few seconds.
If you would like to try this on your own,
pause the video. Otherwise the answer is coming
up in just a few seconds:) Well there you
go. I hope that you got the right answer.
I can't help but talk about this just a tiny
bit. Make sure that are careful when you take
those binomials and you square them. You get
3 terms. Don't try and do it in your head
and forget the middle term and only get 2
terms. A really common mistake from my students
in PreCalculus until I hammer it in to them
enough. This is a parabola. It is smooth and
continuous everywhere in its domain. So the
limit does exist at (2,3) So once again you
will notice that as we simplify our work,
that denominator of delta x does eventually
get cancelled with a factor of delta x that
we can bring out of all of our terms in the
numerator. Thus allow us to just take that
value of 0 that delta x is approaching and
get that slope of 5. I have also displayed
the picture of the parabola here... roughly.
The point of tangency, this purple line is
that tangent line which we just found the
slope of which is 5. Now if you are being
asked to find something more than just the
slope at this particular point of (2,3). To
find the equation of the tangent line, you
have been finding equations of lines since...
what... pre algebra or maybe algebra 1. You
have a point right here of (2,3), you have
a slope of 5, so you can just plug into your
Point Slope form. There is more than one way
to write the equation of a line. But, if you
use point slope form, it is y minus the y
from your point of tangency, your slope which
you just found which is 5, times x minus the
x coordinate of your point of tangent... what
we used as 'c'. We get y-3 is equal to 5x-10.
Add 3 to both sides and we get y is equal
to 5x-7 when you add 3 to both sides. That
is the equation of this tangent line. Ok.
I got one more example. This one is going
to be the most complicated of the three that
we are doing right now where we can actually
find the limit exists at 'c' and such. We
are just finding the slope again at a particular
point. Coming right up! nananananana.... f(x)
is equal to the square root of x-1 at (5,2)
We are going to find the slope of that curve
at that particular point. We are going to
do that with the definition of the derivative.
So we are going to say the slope of our tangent
line is equal to the limit as delta x approaches
zero of the square root of c+Dx... so the
square root of 5+Dx minus 1... minus... now
we are going to do f(c). So we are going to
take the 5 and plug it in. So minus the square
root of 5-1. I don't need to use parenthesis
again because the square root is going to
act as a grouping symbol as well. All over
delta x. That is going to simply, 5-1 which
is 4. The square root of 4 is 2. That is going
to simplify to the limit as Dx approaches
0 of the square root of 5-1 which is 4 minus
2 all over Dx. I said I was going to reveal
the solution and let you refer to those other
videos since some of these examples are repeats.
But if this is the only video you watch, you
are not going to want to go find another one.
So I am going to walk through this. Hopefully
you remember this a little bit as you started
finding your limits with properties of limits.
And we can manipulate this expression... if
not, let's take a look at it. We can manipulate
this expression so that we can actually take
the value of zero and plug it in. Right now
of course, we are going to be trying to divide
by zero and that makes this expression undefined.
So the only other way... If I cannot manipulate
this to get that denominator to be such that
I can plug in zero, I would have to resort
to either finding this limit graphically or
with those tedious tables that we did finding
limits numerically and graphically. So I really
hope that we can manipulate this so that I
can just plug in the zero. And what we are
going to do is, we are going to rationalize
the numerator. So we are going to multiply
the numerator and denominator by the conjugate.
It is going to be the square root of (4+Dx)
+ 2, again in the numerator and denominator.
This is a difference of squares pattern. It
is a-b times a+b. That is factored form of
a difference of squares. So when I distribute
these together my middle term is going to
cancel out. I am going to go ahead and show
that to you. Speed this up a little bit. I
am just going to pause this between the steps.
That was a lot quicker than watching me write
it. I have taken my two binomials and distributed
them together. Of course, again I am pointing
out that difference of squares pattern. These
two middle terms are going to cancel out.
We are also going to have 4 minus 4 of is
zero. So those are going to cancel out. And
I have left my denominator in factored form
if you will. I have not distributed the delta
x because the goal is to get that delta x
to cancel out so I can plug in the value of
zero and get this limit with direct substitution.
Well if all of this cancels out I am just
left with delta x. That one sole remaining
can be divided with itself in the denominator.
And remember this numerator... We are are
not adding and getting zero. We are dividing
and getting the answer of one now. So our
numerator is still going to be 1. It does
not just disappear. Let's get that up here.
Now we can take and plug in our value, our
delta value, let it approach zero. Just do
a direct substitution and it will not come
out to be undefined. Thus we will have the
slope of this curve at that point. That is
going to be 1 over... I am not going to write
the limit notation any more because I am plugging
in zero... over the square root of 4 plus
0... plus 2. That comes out to be 1 over the
square root of 4 is 2, 2+2 is 4. That is the
slope again of my tangent line. Let's keep
my notation neat and do little M sub t so
I know this is the slope of the tangent line.
Now, this is all fine and dandy but.. Oops.
Excuse me. Let me do one more example before
I get into finding the derivative function.
What happens when you try to find the slope
of a smooth continuous curve, but where you
are trying to find the slope of that smooth
and continuous curve... key word on smooth....
What happens if the tangent line is vertical?
That means that we are going to have some
issues trying to get rid of that delta x in
the denominator. So we need to recognize that.
It is coming up right now! Now what is going
to happen with the vertical tangent line.
Don't forget the equation of a vertical line
which by the way is x=# as far as an equation
goes. Vertical lines have a slope which is
undefined. So of course the limit as delta
x approaches zero of our definition here f(c+Dx)-f(x)
over delta x is going to come out to be infinity
or negative infinity. So thus that is why
the last section of the previous chapter was
about infinite limits. We are going to say,
you know, the limit really does not exist.
We might say that it is infinity or negative
infinity, but we are actually by stating that
we are stating that the real limit does not
exist. Thus the slope of our tangent line
is undefined. So over here we are just plugging
along doing our homework and we get this.
Find the slope of a tangent through (0,0)
for f(x) which is equal to x to the 1/5 power.
So the slope here is equal to the limit as
delta x approaches zero and we say... well
ok... it 'c' or 'c'. Excuse me, I should be
looking up here. Excuse me c+Dx to the 1/5
power minus f(x) or 0 to the 1/5 power all
over delta x. Ok, so everything seems good
so far. Then we get the limit as delta x approaches
zero of... well that is just zero, and 0 plus
delta x is just going to be delta x to the
1/5 power. So we end up with Dx^1/5 over Dx.
Even though this is a bit of a different looking
base than maybe you would have had in Algebra
2 honors or Precalculus, it is still a base.
When you divide like bases you take the exponents
and you subtract them. So we have... this
is equal to... Just off to the side here if
you need to, we have delta x to the 1/5 power
minus this exponent of 1. But we need common
denominators, so that 1 is going to become
5/5. That of course is delta x to the -4/5.
So I could say that this is equal to delta
x to the -4/5, or just write in such a way
that I don't have to have that negative exponent.
That means that is is going to be the limit
as delta x approaches zero of 1 over delta
x to the 4/5 power. Really nothing else can
be done to that expression. Yet I need to
let Dx approach zero. If there is no way for
me to manipulate this into a function that
agrees with this one at all points except
'c', then I am left trying to find this limit
by either doing a graphical approach or a
numerical approach. So instead of doing a
long t-table like we did initially when we
found limits numerically, I just want to just
think about this a little bit. If I let x
approach zero from the left, of 1/Dx^4/5 let's
not forget the top number is power and the
bottom number is a root. So if I take a number
that is less than zero because I am approaching
from the left. It is a negative number. I
take a negative number and I take an odd root
to it, it is still going to be negative. But
then I am taking that negative number and
raising it to an even power. Now a negative
number raised to any even power is always
going to come out to be positive. As this
becomes closer and closer to zero, 1 divided
by an infinitely small number becomes infinitely
large. So this is going to approach positive
infinity. And if I do the right hand limit,
well if I plug in a positive number, I take
a fifth root I get a positive answer, I take
that positive answer and I raise it to the
fourth power, it is still positive. I still
have 1 being placed over an infinitely....
now we are approaching zero so again the denominator
is becoming infinitely small approaching zero.
That is going to make the overall fraction
become or approach again positive infinity.
So this limit is, I can say this is a definitive
limit. I can say the limit is infinity because
the left and right limits are both equal.
But I am actually validating that there is
no real limit. The real limit does not exist,
thus we have evidence of a vertical tangent
line. Let me just pause. I want to give you
the graph of this function that we are looking
at and also the graph of... basically what
we have here is 1 over delta x ^4/5... the
derivative here and compare the two. So what
I have done is I have drawn here in purple
roughly what the function f(x)=x^1/5 looks
like. It is a smooth continuous function.
But there is that sharp rise at x=0 and there
is that vertical, well it is actually the
y axis, but our vertical line which is our
tangent line at x=0. In orange, even though
I am saying the definition of slope of the
tangent line as if it is interchangeable with
the definition of the derivative. It is not
exactly correct. So I am getting ahead of
myself here a little bit because of all these
concepts I am trying to tie together for you.
But f prime of x or the derivative of f(x)
is equal to 1/x^4/5. I am just taking this
delta x here, we are still trying to find
the derivative at a point. We can see that
there is some kind of variable here to the
4/5 in the denominator. I wanted to show you
how it approaches infinity, or it approaches
infinity as you approach the x value of 0
from the left and the x value of 0 from the
right thus the two sided limit is the same.
It just happens to be infinity which is nice
to say that it is equal to infinity. But we
are actually validating that the real limit
does not exist. Thus the slope of the tangent
line is undefined. The tangent line is vertical
because we had that limit of either infinity
or negative infinity. We also have a couple
of other special cases where we might find
the... or we are going to try and find the
derivative at a sharp bend. You cannot find
the derivative there as well. Also, finding
the derivative of a function when it is a
say step function, or it is discontinuous
in the sense that it is not removable where
the left limit and right hand limit are not
equal. I am trying to decide if I want to
do those... Yes I think I will. I will do
those two other special cases before we get
to actually using the full blown proper definition
of a derivative which means that we are going
to finding the derivative function as opposed
to just the slope at a specific point. Derivative
at a sharp bend. Remember those absolute value
functions. Inside the absolute value function,
where that expression is able to equal zero
and the function without the absolute value
symbol would cross through having positive
y values into negative values or vise versa,
well the absolute value function tells you
how far a value is on the number line. Basically
it lets us see that it just cancels out the
negatives. So we get these sharp bends. As
soon as it hits where it should be hitting
the x axis, unless there is a vertical shift,
it is going to bounce right off it and make
sharp bend at that one particular value of
x. Well that value of x is going to be negative
three here. Because if I plug in -3 I get
0 inside my absolute value function. So we
are going to try and find the slope of that
function, which is again going to be continuous
along its entire domain. Find the derivative
at that point of x=-3, or the 'c' value equals
-3. So we are going to say the slope of the
tangent line is equal to the limit as delta
x approaches 0 of, and we are going to let
this equal c, so it is going to be the absolute
value of -3 plus delta x plus 3 minus... and
we are going to take that value of c and plug
it in... so minus the absolute value -3+3.
Did I just call that -3? It is positive three
of course. All over delta x. Ok. And maybe
we are not even thinking about that sharp
bend showing up. We are just going through
all these definition of the slope of the tangent
line formula over and over again, and this
happens. We have the limit as delta x approaches
zero of... Well let's see here. We have -3
plus 3 is 0. Those cancel out. So we have
the absolute value of delta x over.. Oh this
is zero again. All of a sudden this becoming...
what looks like extremely simple. And we get
to a situation where we have a denominator,
this delta x in the denominator, and not having
any way of cancelling it out. So again, we
can find the limit both numerically with a
table of values like we did before. We can
just think about it like I am about to do.
Or we could use a graph. We are going to be
thinking about what is happening with the
signs here. If I let... I want to find the
limit when we let delta x approach 0 from
the left of this function. Think about what
the absolute value function does. We have
the exact same expression just sitting alone
in the denominator and sitting in the numerator.
If we remember that again, the absolute value
symbol just makes the negatives disappear
if you will after you simplify what is inside.
We can see that is is delta x divided by delta
x which comes out to be some form of 1. So
if we want to approach 0 from the left, that
means that we are plugging in negative numbers.
So if I plug in a negative number into this
numerator, the absolute value symbol is going
to make it positive. We will have a positive
value up here. I guess we could say if we
want to approach from the left, you could
maybe plug in a -.1 and a positive .1. Maybe
we should just do that. Do a test number.
Let's plug in -.1 because that is kind of
close to 0 and to the left. Plug in a -.1
in the denominator. The absolute value is
going to make that positive. That is going
to come out to be -1. Ok, that is fine as
long as the right hand limit is equal to the
same value, then the two sided limit exists.
So the limit as delta x approaches 0 from
the right. That is.. you can think about it
as since this is going to cancel again and
be one. Or you can use a test value. I am
going to go just to the right of 0 and use
positive .1. The absolute value of positive
.1 is going to keep it as .1. That is equal
to one. So we know of course that for a limit
to exist, for the limit as delta x approaches
zero to exist for this function, the left
and right hand limits would have to be the
same. They are not, thus the limit which we
are tying to find the slope of the tangent
does not exist, we cannot find the slope of
a function... we cannot find the limit as
delta x approaches zero... we cannot find
the slope of that tangent line at that sharp
bend. In this case that is when x=-3. We are
also not going to be able to find... the derivative
function is going to be undefined also at
that point since we cannot find the limit.
So we can't find the derivative... you cannot
find the slope of a tangent line at a sharp
because the left and right hand limits are
not equal! I am sorry I'm now using this word
differentiability and derivative, but I just
thought the video would flow better if I did
all these special cases tying to find the
slope of the tangent line at a specific point
before I got into the proper definition of
derivative which is very very similar. Again
here, this was the second example with the
vertical tangent line example I just did and
this one here with the sharp bend. Again f
is continuous at c, but not differentiable
at c, or in other words we cannot find a real
slope of the tangent line. However if f is
differentiable at c, then f is continuous
at c. Or in other words differentiability
implies continuity, we will get to that proper
definition of finding derivatives here after
one more example. So if you can find a derivative
at a point, if you can find the real slope
of the tangent line at a point then the function
is continuous at that point. But, we have
seen that just because it is continuous does
not mean that we can find the derivative or
the slope of that tangent line at that point.
I have got one more example. This is one is
going to be a floor function or a greatest
integer function just to finish our last special
case before we get into the proper definition
of derivative and finding derivative function
as opposed to just slopes at given point.
nanananana.... Now this is coolest part about
this section, amongst all of these other examples
I am giving you of course so you can do your
homework. Here we are going to take the difference
quotient and we are going to take out the
value of 'c' and replace it with x. So instead
of finding the slope at one fixed point on
a curve, that is find... but if we are finding
the slope at x equals 2 and I want to find
the slope at x=4, I have to go through this
whole limit process again and you can see
that it is not very quick. So if we replace
that c with x, that means that we are going
to do the same process, but in general terms
as opposed to at one specific value of c.
What that is going to do for us is find a
derivative. This is the definition of the
derivative that... I was incorrectly saying
definition of derivative when we were actually
doing the definition of the slope of the tangent
line at a point. This is much more power in
the sense that when I do this process I get
something called f prime of x, or the derivative
of f. It is going to be.. Well let's see.
It is going to be another function in terms
of x and that function is going to tell us
the slope of the original function at any
point. So when I walk through this limit process
one time, the original time, we this new function...
this derivative function. Then if I want to
find the slope of the original f(x) at 2 and
then at 4 all I have to do is plug 2 into
this function. I get the slope of the original
function f(x). Plug 4 into this function and
assuming that the limit exists and I will
have my second slope. So I can find the slope
at any point along this curve I like as long
as this limit exists at that point. So this
is for all x in which the limit exists. f
prime is a function of x. f'(x) gives the
slope of the tangent line to the graph of
f(x) as long as f(x) has a tangent line at
that point (x,f(x)). We already talked about
how difficult it is... or the issues that
we have with vertical tangent lines having
this limit be undefined or actually equal
to infinity or negative infinity. We tried
to find the slope at a sharp bend, so we can't
always find the slope of curve at ANY point,
but almost all of them. Differentiation is
called the process of finding the derivative.
So I am calling f prime of x, but I can also
say that this is the derivative of f(x). A
function is differentiable at x if its derivative
exists at x. We have had a couple of examples
already where we had an issue with that like
I just said. It is differentiable on the open
interval of (a,b) if it is differentiable
at every single point along that open interval
of (a,b) f'(x), as you can hear, I am reading
f prime of x is f'(x), or the derivative of
y with respect to x. Those are interchangeable.
A lot of times just to ease the notation and
to write a little bit of short hand we call
dy/dx y prime. Or we can use d/dx of f(x).
Then we have this notation which honestly
I don't use and I am not going to be using
in my notes. My textbook mentions it, D sub
x of y. See, I am even pausing on how to say
that because I don't use that notation. This
is again read the derivative of y with respect
to x. This d over dx, this is like the process
of... I am going to find the derivative with
respect to x of my function f(x). Finding
the derivative of f(x) with respect to x.
Now we can also find with respect to 't' a
lot of times if we are doing any kind of time
plot doing some kind of problem that has a
function in terms of time. We can find the
derivative in terms of time. All kinds of
things. Of course we are just getting started
with derivatives, so how about we just keep
it with respect to x. Now I have got two more
examples. I know that we are already at 40
minutes. Thank you for watching and still
hanging on. I am also going to at the end
of this video be doing again that greatest
integer function where the function is discontinuous
at some point where we would like to find
the derivative at. Attempt to show you that
you actually cannot find the derivative because
it is discontinuous. And we are going to be
finding the equation of a tangent line that
is parallel to a given line, and also find
the equation of a tangent line that goes through
a point that is not on the curve. Hopefully
this one video helps you get through all your
homework probably at the beginning of learning
of course about your derivatives with that
definition. Find the derivative by the limit
process. We have f(x)=2x^2-3x+1. This is a
parabola. We know parabolas to be smooth and
continuous along their entire domain of all
real numbers. So no problem with this having
any issues with being able to find the derivative.
It is going to be differentiable along all
real numbers from negative infinity to positive
infinity. So f'(x) or the derivative of f
is going to be equal to the limit as delta
x approaches 0... now remember we not plugging
in c plus delta x, we are plugging in x plus
delta x. It is going to be in general terms.
So 2 times x+Dx squared minus 3 times x+Dx
plus 1 minus... now that we are actually going
to find the derivative. Sort of like the general
form of the slope, we are not going to plugging
in a value of c which just gives us a constant
for this polynomial. That means.... Here is
where I was trying to emphasis that it is
important for you, with that minus sign and
subtracting here, to use parenthesis. Let
me just write this out here first. 2x^2-3x+1.
Remember to wrap that functions in parenthesis
so that you have to carry that negative sign
through the expression so that you don't make
any sign errors. The rest of this problem
is going to be extremely similar in the mathematical
steps of my previous examples. So i am going
to step off. You might want to pause the video
and try to finish this out for yourself and
see what the derivative is. I am not going
to walk through this step by step because
of all the examples I have already done. The
Algebra is the same, just a lot more variables
because this does not come out to be a constant
value. Of course we don't have any constants
in here as well. So just be more careful with
your algebra, do it one step at time, don't
do too many things at once or anything in
your head so that if you do make a small mistake
you can find it and correct your work. Alrighty
then. Here is all the algebra. Rest assured
by the way in the upcoming sections we are
going to learn a quicker way of finding these
derivatives. You might also as we go through
these couple of examples, or your homework,
look at your original function and your derivative
and start seeing if you notice a relationship
between the two. That might give you a heads
up on how we are going to be approaching these
derivatives in the upcoming sections. Keep
a note of this derivative, the slope of this
function. I am going to be using that in my
last two examples in this video. Alright.
I have another one coming up. It is a fraction
and it involves radicals. So guess what? It
will be kind of hard. [laughing] We will make
it easy. Same directions. Find the derivative
by the limit process. So f'(x), or the derivative
of f is equal to the limit as delta x approaches
zero... Now remember we are plugging in again
x plus delta x. So we have 2 over the square
root of x+Dx minus 2 over the square root
of x all over Dx. So we are going to have
quite a bit of work of finding common denominators
and working with fractions which we all know
we love to do! So we are going to find a common
denominator. We are going to do that by multiplying
the first fraction by square root of x in
the numerator and denominator and the second
fraction by the square root of x+Dx. And to
not force you to watch me write, we will just
make it appear:) Well you have a fraction
divided by a fraction. So hopefully you remember
that when you divide fractions you need to
multiply by the reciprocal of the second fraction.
That second fraction is the denominator. So
we are going to take this delta x in the denominator
and move it up with multiplication. We have
1 over delta x. Now don't worry about taking
all these factors in the denominator and multiplying
them together. Remember we are trying to get
that delta x to go away so I can plug in zero.
If I can make it go away. We have a binomial
in the numerator with 2 times the square root
of x times 2 times the square root of x+Dx.
And just like learned and should have been
practicing when you learned about finding
limits with properties of limits and even
earlier in the video, we want to multiply
the numerator and denominator by the conjugate
of this binomial to get these square roots
to cancel out. That is not the main purpose
of course. Hopefully we manipulate this again
so we can let that delta x go to zero by getting
it to cancel with something in the numerator.
So let's see what happens. We have multiplied
by that conjugate which of course is that
a-b times a+b pattern, the difference of squares
pattern when you are factoring. I know that
I am not going to have that middle term. Usually
you get a trinomial when you multiply the
two binomials. So we have 2 times 2 which
is 4. The square roots cancel out on the x.
Same thing happens over here, -2 times positive
2 is negative 4 times... again that radical
cancels out we get x+Dx. Now don't forget
to wrap that in parenthesis because square
root symbols are not only a math function
but it is an actual grouping symbol as well.
So if you don't wrap that x+Dx in parenthesis,
you will forget to take this -4 and distribute
it through the parenthesis and have sign errors
and probably think that this delta x does
not cancel out, which it will. So we have
4x-4x is equal to of course 0. And now that
we have one term in the numerator -4Dx, and
that Dx is now a factor... not just part of
a term, we can cancel out this Dx from the
negative four leaving -4 as my numerator.
Right there. That looks a little confusing,
like it might be negative eight. So let's
just circle that lone remaining -4. Now I
can take, with that delta x factor out of
my denominator, take that 0... plug it in...
a lot of nice things are going to happen and
we will be done. f'(x) is equal to -4 over...
now let's not forget we are plugging in zero,
right. We are letting the delta x approach
zero. So this is going to approach zero, this
delta x is going to approach zero, and we
have the square root of x times... now this
is just going to be the square root of x...
so square root of x times square root of x
is x. We have got 2 square root of x plus
2 times... oh well I am adding by zero so
that is going to be basically 2 square root
of x again. So 2 square root of x plus 2 square
root of x is 4 square root of x. Now again
my numerator and denominator have one more
factor that can cancel out, that is that common
factor of 4. My final answer is the derivative
of f, or f prime of x, is equal to -1 over
x times the square root of x. Ok, let's take
a look at trying to find the derivative with
that greatest integer function and do those
last two examples. So now we are going to
take a look at trying to find the derivative
at a point which the function is not continuous.
What happens? Well it turns out to be not
differentiable. If f is not continuous at
c, so we are going back to finding a derivative
at a point instead of just the derivative
function, is also not differentiable at x=c.
In our example we are going to use the greatest
integer function which gives us values of
n such that n is going to be less than or
equal to x. Now to work through this example
with these tables I am going to use an alternative
form of the derivative which is f'(c) is equal
to the limit as x approaches c of f(x)-f(c)
over x-c. It kind of looks like your.... It
does look exactly like your average rate of
change formula from Precalculus. But, again
the limit process is going to allow the value
of x approach c thus taking this point and
pushing it closer and closer and closer to
our original fixed point we want to find the
slope at which is (c,f(c)) thus giving us
the slope of the tangent. Well what happens?
It looks like.. If I look at this graph here,
this line is horizontal. Remember the derivative
is slope. So our slope here is zero. Our slope
here seems to be zero, and zero, and zero.
But obviously at the value of 1, 2, or 3...
we are going to let c equal 1 by the way and
try and find the derivative at x=1... After
I get that phone. It has already been picked
up. We have to jump from a y value of zero
to a y value of 1. So clearly something is
happening there and we are going to have an
issue trying to find the derivative. There
is no easy way to get rid of the x-1 in the
denominator so we are going to find the left
and right limit using table just like we did
in the first section about finding limits
and finding them numerically and graphically.
We are going to let x equal .9. If I plug
.9 into my greatest integer function it is
going to give me zero, and zero minus one
is -1 over .9-1. You simplify that and you
get a value of 10. Move a little bit closer
to 1 from the left and you get an answer of
100. Move a little closer and you get 1000.
Move a little closer and you are going to
approach a slope or a left limit of this derivative
function which infinity. So yes, as you approach
1 from the left the slope is going to go to
infinity to get up here to the next step of
our greatest integer function. Now as you
approach from the right.... I have already
shown with the one sided limit that it is
infinity. Now we talked about infinite limits,
but stating the limit is infinity is stating
a real limit does not exist. As you approach
from the right and we take 1.001, plug it
in here and see what the value is, well 1.001
plugged into the greatest integer function
brings it down to 1, and 1-1 is equal to 0.
Plug in 1.01. 1.01 plugged into our greatest
integer function comes out to be 1. 1-1 again
is equal to 0 over something but if the numerator
is equal to zero then all of these values
are going to be zero. So as I approach 1 from
the right this limit is equal to zero. Also
remember, you might glance at this and go...
Well I am approaching 1 from the right. The
limit of this function as I come in from the
right of 1 is equal to 1. But remember we
are trying to find the derivative which is
slope and this is a horizontal line. It kind
of makes sense then that this horizontal line
would have a slope equal to zero. But again
on the left hand side approaching 1 with that
undefined value and then you had to step up
to the next step of our greatest integer function,
the left and right limits do not match therefore
the limit does not exist. Therefore nor does
the derivative. The function is not differentiable
at that point of c where the function is not
continuous. BAM!!! Find the equation of the
tangent line to the graph of f(x)=2x^2-3x+1
that is parallel to y-x-7=0. So I want to
find the slope of the line that is parallel
to that line. You have been doing that since
algebra 1. You need a point and you need a
slope. Well let's see here. What is the slope
of the given line? If I take this and I solve
it for y, I get y=x+7. So the line that I
am given has a slope of 1. I want to have
a parallel line so I want to use that slope
of 1. Ok, well what is the slope of the function
because this is a parabola. We know that it
is constantly curving. We can find that slope.
Where is the slope of this parabola equal
to 1. Are we going to be able to find that?
I can do the definition of the slope of a
tangent line at a point. But I could just
write that derivative down like I said we
were going to use in that previous example.
The derivative of f(x) is equal to 4x-3. If
I know the derivative of this function, the
derivative gives me the slope of this function
any point where the limit exists... or the
derivative exists... I can say I need to have
a slope of 1. That means my derivative, my
slope, of that parabola must equal 1. So 1
equals 4x-3. When I solve that for x I am
going to add both sides by 3 and get 4 is
equal to 4x. Divide both sides by 4 and get
1=x. Well again you need a point and a slope
to write the equation of a line. I have got
that slope once again. But now I only have
half of the point. I need to take that value
of x=1 and plug it back into the derivative
which is going to give us a slope, but back
into the original function so I know that
the point of tangency between this parabola
and the tangent line I am about to find the
equation of. The point of tangency is f(1)
is equal to 2(1)^2-3(1)+1. That is going to
come out to be 2 times 1 squared is 2, 2 minus
3 is -1, now plus 1 is equal to zero. So my
point of tangency is (1,0). Now I have a point
and I have a slope. Obviously I have made
these numbers to work out nicely. The tangent
line is equal to y minus the y of my point
of tangency which is... I am using point slope
form... y minus 0 is equal to 1 times x minus
the x of my tangency point. That works out
to be y=x-1. Let me just double check. YES!
Alright. If you are still watching, thank
you very much. I have got one more example
and we will be done. The last example in our
video is going to be probably be the one that
requires the most thought from you to figure
out how to do these problems. Find the equation
of a tangent line to f(x)... again our common
parabola here I just got doing the derivative
of... I want a tangent line that touches...
of course it tangent to that parabola and
goes through this point. Now we have a parabola,
I just did a quick sketch of what is going
on. We have this point of (3,8) over here
to the right in space. We are going to end
up having two tangent lines that will go through
that point and touch the parabola in only
one point, the point of tangency. So of course
we need a point and a slope to find the equations
of lines. So lets start off with slope. The
slope of the function is the derivative. And
f'(x) is equal to 4x-3. Ok, well I need another
way of looking at the slope so that I can
incorporate this point of course and get two
answers. We have a y and x... and that is
not enough. I need another way of writing
slope. I need a... This is the slope of the
parabola that is constantly changing but between
that point of tangency and this coordinate
that is out in space, I have to use the (3,8)
somehow... I need to write a slope formula
for any point on the function and this exterior
point of you will that is not on the parabola
of (3,8). So slope we know is equal to Y2-Y1
over X2-X1. I am going to let this point here,
this (3,8) be my X1 and Y1. So I am going
to put a 3.... Excuse me that is supposed
to be an x. I am going to put a 3 here and
an 8 there. But I need a better way of writing
Y2 and X2. How do we find the Y? Where is
that going to come from? It is a y coordinate
that is on this parabola. How do you find
y values? Like graphing with t-tables, you
plugged in an x and found a y. So I am going
to rewrite this replacing Y2 with... you might
have it written as f(x)... with this function.
Alright. I have two ways of looking at slope
now. I have got the derivative which is 4x-3
and I have a slope formula that actually incorporates
the value of (3,8) right here. Now what I
want to do is put these together so I can
solve for x and find our points... or at least
our x coordinates... for our two points of
tangency. That means I am going to take out
the m... I am just using the Substitution
Method to solve a system of equations like
you did at the end of Algebra 1 and Algebra
2. So the 4x-3 goes here. We are going to
work this out. We are going to multiply both
sides of our equation... Let's just wrap that
up in a big parenthesis... with x-3. This
is going to cancel out that denominator. When
we distribute these together we are going
to get 4x^2-3x-12x... -3 times -3 is 9. That
is going to be 4x^2-15x-9 is equal to... Over
here this has a power of 2, a power of 1,
and we have got 1-8 which is going to be -7.
We are going to have a quadratic here coming
up. We are going to have to get this set equal
to 0. I am going to bring the 2x^2 over with
subtraction. We are going to bring the -3x
over with addition. We are of course going
to bring the -7 over with addition as well
doing the inverse math operations. We get
2x^2-12x+16 is equal to zero. We are going
to solve this for x. We are going to take
the 2 out first. This is a factorable trinomial.
We are looking for factors of 8 that add to
-6. That is going to be -4 and -2. So our
points of tangency are at x equals 4 and x
equals 2. Well I am going to repeat... This
is kind of like the same thing now as I finished
my last problem. I will have a slope. But
the first thing I need to do is look at these
two points of tangency. I need my two points
of tangency's that come from having an x coordinate
of 4 and an x coordinate of 2. So I am going
to get this board erased, to a little bit
of labelling, and I will be right back. So
I went ahead and marked the graph. I have
found our points of tangency by taking our
two values of 2 and 4 and plugged them back
into our original function. Now we have got
two points. That means that we are going to
have two of those tangent lines again. We
have slope. Now I didn't the slope out yet
for each of these lines. But the derivative
is going to tell us the slope at the x coordinate
of 2 and the x coordinate of 4 as I work out
this problem instantaneously. I have got y
minus... I am going to use Point-Slope form.
y minus the y of our point which is 3 is equal
to m... Ok. Well that is going to tell us
what m is. 4 times our x coordinate is equal
to 2 minus 3. So that is our y minus Y1 is
equal to M times x minus X1. 4 times 2 is
8, and 8 minus 3 is 5. Take the 5 and distribute
it through the parenthesis and I am going
to get y-3 is equal to 5x-10. Add both sides
by 3 and one of our tangent lines is y=5x-7.
Negative 10 plus 3 is equal to -7. The other
tangent line is going to be y minus the y
coordinate which is 21 is equal to 4 times
the x coordinate... we are find slope of the
function at 4... So again 4(4)-3 now times
x minus the x coordinate here of 4. Of course
with the same process of simplifying this
algebra we are going to get.... Let me just
double check my work here. Yes we are going
to get a final answer of y is equal to 13x-31.
If you got though this very long video, thank
you very much. I hope this matches the problems
that are in your textbook where you are dealing
with the definition of the derivative and
the tangent line problem. I am Mr. Tarrou
and I am finally out of here! BAM! Go Do Your
Homework:)
